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BY THE SAME AUTHOR.
A Treatise on Elementary Dynamics. Crown 8vo.
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PART I. ELEMENTS OF STATICS.
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|
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aontion: C. J. CLAY and SONS,
CAMBRIDGE UNIVERSITY PRESS WAREHOUSE,
AVE MARIA LANE.
(ffilagfloia: 263, ARGTLE STREET.
THE ELEMENTS
OP
COOEDINATE GEOMETRY.
THE ELEMENTS
OF
COOEDINATE aEOMETRY
BY
S. L. LONEY, M.A.,
LATE FELLOW OF SIDNEY SUSSEX COLLEGE, CAMBRIDGE,
PROFESSOR AT THE ROYAL HOLLOWAY COLLEGE.
^^^l^^mU. MASS.
MATH, DEPTi
MACMILLAN AND CO.
AND NEW YOEK.
[All Bights reserved.']
CTambrtlJse:
PRINTED BY J. & C. F. CLAY,
AT THE UNIVEBSITY PRESS.
PKEFACE.
"TN the following work I have tried to present the
elements of Coordinate Geometry in a manner
suitable for Beginners and Junior Students. The
present book only deals with Cartesian and Polar
Coordinates. Within these limits I venture to hope
that the book is fairly complete, and that no proposi-
tions of very great importance have been omitted.
The Straight Line and Circle have been treated
more fully than the other portions of the subject,
since it is generally in the elementary conceptions
that beginners find great difficulties.
There are a large number of Examples, over 1100
in all, and they are, in general, of an elementary
character. The examples are especially numerous in
the earlier parts of the book.
vi PREFACE.
I am much indebted to several friends for reading
portions of the proof sheets, but especially to Mr W.
J. Dobbs, M.A. who has kindly read the whole of the
book and made many valuable suggestions.
For any criticisms, suggestions, or corrections, I
shall be grateful.
S. L. LONEY.
EoTAIi HOLLOWAY COLLEGE,
Egham, Surbey.
July 4, 1895.
CONTENTS.
CHAP. PAGE
I. Introduction. Algebraic Kesults ... 1
II. Coordinates. Lengths of Straight Lines and
Areas of Triangles 8
Polar Coordinates 19
III. Locus. Equation to a Locus 24
IV. The Straight Line. Eect angular Coordinates . 31
Straight line through two points .... 39
Angle between two given straight lines . . 42
Conditions that they may be parallel and per-
pendicular . . . . . . .44
Length of a perpendicular . . " . . 51
Bisectors of angles 58
V The Straight Line. Polar Equations and
Oblique Coordinates . . . . 66
Equations involving an arbitrary constant . . 73
Examples of loci 80
VI. Equations representing two or more Straight
Lines 88
Angle between two lines given by one equation 90
Greneral equation of the second degree . . 94
VII. Transformation of Coordinates . . . 109
Invariants 115
Vlii CONTENTS.
CHAP. PAGE
VIII. The Circle 118
Equation to a tangent 126
Pole and polar 137
Equation to a circle in polar coordinates . .145
Equation referred to oblique axes . . . 148
Equations in terms of one variable . . .150
IX. Systems of Circles 160
Orthogonal circles . . , . . . .160
Kadical axis 161
Coaxal circles 166
X. Conic Sections. The Parabola . 174
Equation to a tangent 180
Some properties of the parabola . . . 187
Pole and polar 190
Diameters 195
Equations in terms of one variable . . .198
XI. The Parabola {continued') .... 206
Loci connected with the parabola . . . 206
Three normals passing through a given point . 211
Parabola referred to two tangents as axes . .217
XII. The Ellipse 225
Auxiliary circle and eccentric angle . . .231
Equation to a tangent . . . . . 237
Some properties of the ellipse .... 242
Pole and polar 249
Conjugate diameters ...... 254
Pour normals through any point . . . 265
Examples of loci 266
XIII. The Hyperbola 271
Asymptotes 284
Equation referred to the asymptotes as axes . 296
One variable. Examples 299
CONTENTS. IX
CHAP. PAGE
XIV. Polar Equation to, a Conic .... 306
Polar equation to a tangent, polar, and normal , 313
XV. General Equation. Tracing of Curves . 322
Particular cases of conic sections .... 322
|
Transformation of equation to centre as origin 326
Equation to asymptotes 329
Tracing a parabola ...... 332
Tracing a central conic . . . . . . 338
Eccentricity and foci of general conic . 342
XVI. General Equation ...... 349
Tangent 349
Conjugate diameters ...... 352
Conies through the intersections of two conies . 356
The equation S=Xuv 358
General equation to the j)air of tangents drawn
from any point ...... 364
The director circle ....... 365
The foci 367
The axes 369
Lengths of straight lines drawn in given directions
to meet the conic 370
Conies passing through four 23oints . . . 378
Conies touching four lines 380
■ The conic LM=B? 382
XVII. Miscellaneous Propositions .... 385
On the four normals from any point to a central
conic 385
Confocal conies ....... 392
Circles of curvature and contact of the third order . 398
Envelopes 407
Answers . i - xiii
ERKATA.
Page 87, Ex. 27, line 4. For "JR" read " S."
„ 235, Ex. 18, line 3. For "odd" read "even."
,, „ ,, ,, line 5. Dele "and Page 37, Ex. 15."
,, 282, Ex. 3. For "transverse" read "conjugate."
CHAPTER I.
INTRODUCTION.
SOME ALGEBRAIC RESULTS.
1. Quadratic Equations. The roots of the quad-
ratic equation
a'3^ + 6x + c = 0
may easily be shewn to be
- & + •JlP' - 4ac 1 -b- s/b^ - 4:aG
They are therefore real and unequal, equal, or imaginary,
according as the quantity b^ - iac is positive, zero, or negative,
i.e. according as b^ = 4:ac.
2. Relations between the roots of any algebraic equation
and the coejicients of the terms of the equation.
If any equation be written so that the coefficient of the
highest term is unity, it is shewn in any treatise on Algebra
that
(1) the sum of the roots is equal to the coefficient of
the second term with its sign changed,
(2) the sum of the products of the roots, taken two
at a time, is equal to the coefficient of the third term,
(3) the sum of their products, taken three at a time,
is equal to the coefficient of the fourth term with its sign
changed,
and so on.
L. e 1
COORDINATE GEOMETRY.
Ex. 1. If a and /3 be the roots of the equation
b c
a a
we have
a + p= - and a^ = -
Ex. 2. If a, j8, and 7 be the roots of the cubic equation
ax^ + bx^ + cx + d=0,
i.e. of
we have
and
o-Pl-
3. It can easily be shewn that the solution of the
equations
a^x + h^y + G^z = 0,
and a^ + h^y + c^z = 0,
IS
X
y
^1^2 ~ ^2^1 ^1^2 ~ ^2^1 '^1^2 ~ ^2^1
Determinant Notation.
4. The quantity-
is called a determinant of the
second order and stands for the quantity a-})^ - aj)^, so that
d-yf d^
^1, h
Exs. (1) ;' | = 2x5-4x3 = 10-12=-2;
!4, 5i
(ii)
= - 3 X ( - 6) - { - 7) X ( - 4) = 18 - 28 = - 10.
DETERMINANTS.
5. The quantity
»2J
«3
&2J
Cl,
^2 5
^3
is called a determinant of the third order and stands for the
quantity
a. X
^2 J ^3
^2 5 <^3
a> *^3i
61,62
i.e, by Art. 4, for the quantity
«i (^2^3 - ^3^2) -- «^2 (^1^3 - &3C1) + ^3 (^i^^a - ^2^1)*
i.e. % (62C3 - h..G^ + (^2 (63C1 - 61C3) + «3 (61C2 - 62C1).
6. A determinant of the third order is therefore reduced
to three determinants of the second order by the following
rule :
Take in order the quantities which occur in the first row
of the determinant ; multiply each of these in turn by the
determinant which is obtained by erasing the row and
column to which it belongs ; prefix the sign + and - al-
ternately to the products thus obtained and add the
results.
Thus, if in (1) we omit the row and column to which a^
belongs, we have left the determinant ^'
^ i and this is the
coefficient of a-^ in (2).
Similarly, if in (1) we omit the row and column to which
a^ belongs, we have left the determinant ^'
and this
with the - sign prefixed is the coefficient of a^ in (2).
7. Ex.
The determinant
1,
5,-6
8, -9
X
5,-6
8,-9
■3)x
-4,5
-7,8
= {5x(-9)-8x(-6)}+2x{(-4)(-9)-(-7)(-6)}
-3x{(-4)x8-(-7)x5}
= {-45 + 48} +2(36-42} -3 {-32 + 35}
= 3-12-9= -18.
1 - 2
COORDINATE GEOMETRY.
8. The quantity
(h.1 ^2> %J ^4
61, &2) hi h
^11 ^25 ^3>
|
j ^1) ^2 5 ^3) ^4
is called a determinant of the fourth order and stands for
the quantity
«i X
K h, ^4
^2» ^3 J
i^lJ ^35 h
- Clo X \ C-,
+ 6^3 X
1 1 5 3 3 4
&i, 62J ^4!
C^ cCj_ X
1 ? 2 5 4
Cl,
^2) Cg
c?i,
»2J <^3
and its value may be obtained by finding the value of each
of these four determinants by the rule of Art. 6.
The rule for finding the value of a determinant of the
fourth order in terms of determinants of the third order is
clearly the same as that for one of the third order given in
Art. 6.
Similarly for determinants of higher orders.
9. A determinant of the second order has two terms.
One of the third order has 3x2, i.e. 6, terms. One of the
fourth order has 4 x 3 x 2, -i.e. 24, terms, and so on.
10. Exs.
2, -3
4, 8
Prove that
= 28. (2)
9, 8, 7j
6, 5, 4 =0.
3, 2, l|
a, h, g
-9
= 85
5,
9,
-3, 7
4,-8
3, -10
-a, b, c
a, -b, c
=:4a6c.
a, I
-98.
9, f, c
= abc + 2fgh - ap - bg^ - ch\
ELIMINATION. 5
Elimination.
11. Suppose we have the two equations
aj^x + a^y = 0 (1),
\x +b^y ^0 (2),
between the two unknown quantities x and y. There must
be some relation holding between the four coefficients 6*i, ctaj
bi, and 63 • ^or, from (1), we have
and, from (2), we have - = - =-^ . K
X
Equating these two values of - we have
i.e. a-J)^ - ajb^ = 0 (3).
The result (3) is the condition that both the equations
(1) and (2) should be true for the same values of x and y.
The process of finding this condition is called the elimi-
nating of X and y from the equations (1) and (2), and the
result (3) is often called the eliminant of (1) and (2).
Using the notation of Art. 4, the result (3) may be
This result is obtained from (1) and (2) by taking the
coefficients of x and y in the order in which they occur in
the equations, placing them in this order to form a determi-
nant, and equating it to zero.
written in the form
0.
12. Suppose, again, that we have the three equations
\x+ h^y^ h^z = 0 (2),
and G^x + G^y + C3S = 0 (3),
between the three unknown quantities x, y, and z.
COORDINATE GEOMETRY.
By dividing each equation by z we have three equations
X
y
between the two unknown quantities - and -
z z
Two of
\y
h
Ci,
^2 1
Cs
these will be sufficient to determine these quantities. By
substituting their values in the third equation we shall
obtain a relation between the nine coefficients.
Or we may proceed thus. From the equations (2) and
(3) we have
Substituting these values in (1), we have
«1 (^2^3 - ^3^2) + «2 (^3^1 - ^1^3) + «3 (^1^2 - ^2^1) = 0. . .(4).
This is the result of eliminating cc, 3/, and % from the
equations (1), (2), and (3).
But, by Art. 5, equation (4) may be written in the form
This eliminant may be written down as in the last
article, viz. by taking the coefficients of x, y, and z in the
order in which they occur in the equations (1), (2), and (3),
placing them to form a determinant, and equating it to
zero.
13. Ex. What is the value of a so that the equations
ax + 2y + 3z = 0, 2x-3y + 4:Z = 0,
and 5x + 7y-8z=:0
may be simultaneously true ?
Eliminating x, y, and z, we have
a, 2, 3,
2, -3, 41 = 0,
5, 7, -8!
^.e. « [( - 3) ( - 8) - 4 X 7] - 2 [2 X { - 8) - 4 X 5] + 3 [2 X 7 - 5 X ( - 3)]=0,
i.e. «[-4]-2[-36] + 3 = 0,
^, ^ 72 + 87 159
4 4
ELIMINATION.
14. If again we have the four equations
a-^x + dil/ + cf'zZ + a^u = 0,
h^x + h^y + b^z + b^u = 0,
Ci«; + c^i/ + G^z + c^u = 0,
and djX + d^y + d.^z + d^ - 0,
it could be shewn that the result of eliminating the four
quantities cc, y, z^ and u is the determinant
«1J
«4
bz,
Ci,
^it
C4
c?i,
C?2,
C?3,
A similar theorem could be shewn to be true for n
equations of the first degree, such as the above, between
n unknown quantities.
It will be noted that the right-hand member of each of
the above equations is zero.
CHAPTER II.
COORDINATES. LENGTHS OF STRAIGHT LINES AND
AREAS OF TRIANGLES.
|
15. Coordinates. Let OX and 07 be two fixed
straight lines in the plane of the paper. The line OX is
called the axis of cc, the line OY the axis of y, whilst the
two together are called the axes of coordinates.
The point 0 is called the origin of coordinates or, more
shortly, the origin.
From any point F in the
plane draw a straight line
parallel to OF to meet OX
in M.
The distance OM is called
the Abscissa, and the distance
MP the Ordinate of the point
P, whilst the abscissa and the
ordinate together are called
its Coordinates.
Distances measured parallel to OX are called a?, with
or without a suffix, {e.g.Xj, x.-^... x\ x",...), and distances
measured parallel to OY are called y, with or without a
suffix, (e.g. 2/i, 2/2,--- 2/'. y",---)-
If the distances OM and MP be respectively x and ?/,
the coordinates of P are, for brevity, denoted by the symbol
Conversely, when we are given that the coordinates of
a point P are (x, y) we know its position. For from 0 we
have only to measure a distance OM { - x) along OX and
COORDINATES. 9
then from 21 measure a distance MP {=y) parallel to OY
and we arrive at the position of the point P. For example
in the figure, if OM be equal to the unit of length and
MP= WM, then P is the point (1, 2).
16. Produce XO backwards to form the line OX' and
YO backwards to become OY'. In Analytical Geometry
we have the same rule as to signs that the student has
already met with in Trigonometry.
Lines measured parallel to OX are positive whilst those
measured parallel to OX' are negative ; lines measured
parallel to OY are positive and those parallel to OY' are
negative.
If P2 b® i^ *li® quadrant YOX' and P^M^, drawn
parallel to the axis of y, meet OX' in M^^ and if the
numerical values of the quantities OM^ and J/aPg be a
and h, the coordinates of P are {-a and h) and the position
of Pg is given by the symbol ( - a, h).
Similarly, if P3 be in the third quadrant X'OY', both of
its coordinates are negative, and, if the numerical lengths
of Oi/3 and J/3P3 be c and d, then P3 is denoted by the
symbol ( - c, - d).
Finally, if P4 lie in the fourth quadrant its abscissa is
positive and its ordinate is negative.
17. Ex. Lay down on "paper the position of the points
(i) (2, -1), (ii) (-3, 2), and (iii) (-2, -3).
To get the first point we measure a distance 2 along OX and then
a distance 1 parallel to OF'; we thus arrive at the required point.
To get the second point, we measure a distance 3 along OX', and
then 2 parallel to OY.
To get the third point, we measure 2 along OX' and then
3 parallel to OT.
These three points are respectively the points P4 , P., , and Pg in
the figure of Art. 15.
18. When the axes of coordinates are as in the figure
of Art. 15, not at right angles, they are said to be Oblique
Axes, and the angle between their two positive directions
OX and 07, i.e. the angle XOY, is generally denoted by
the Greek letter w.
10 COORDINATE GEOMETRY.
In general, it is however found to be more convenient to
take the axes OX and OZat right angles. They are then
said to be Rectangular Axes.
It may always be assumed throughout this book that
the axes are rectangular unless it is otherwise stated.
19. The system of coordinates spoken of in the last
few articles is known as the Cartesian System of Coordi-
nates. It is so called because this system was first intro-
duced by the philosopher Des Cartes. There are other
systems of coordinates in use, but the Cartesian system is
by far the most important.
20. To find the distance between two points whose co-
ordinates are given.
Let Pi and P^ be the two
given points, and let their co-
ordinates be respectively {x^ , y^)
and (a^sj 2/2)-
Draw Pji/i and P^M^ pa-
rallel to OY, to meet OX in
J/j and M^. Draw P^R parallel
to OX to meet M-^P^ in R. q ' M jvT
Then
P^R = M^Mt^ = OM^ - OMc^ = oi^-X2,
RP, = M,P,-M,P, = y,~y,,
and z P^i^Pi = z6>ifiPa-l 80° -PiJfiX^l 80° -<o.
We therefore have [Trigonometry, Art. 164]
|
P^P^^ = P^R^ + RP^^ - 2P^R . PPi cos P^RP^
- (^1 - x^Y + (2/1 - 2/2)' - 2 (a^i - x^) (2/1 - 2/2) cos (180° - (o)
= (Xi-X2)2 + (yj_y2)2+2(Xi-X2)(yi-y2)COSCO...(l).
If the axes be, as is generally the case, at right angles,
we have <o == 90° and hence cos to = 0.
The formula (1) then becomes
DISTANCE BETWEEN TWO POINTS.
SO that in rectangular coordinates the distance between the
two points (x^j y^ and (a-g, 2/2) is
Cor. The distance of the point (x^, y-^ from the origin
is Jx^ + 2/1^, the axes being rectangular. This follows from
(2) by making both x^ and y^ equal to zero.
21. The formula of the previous article has been proved for the
case when the coordinates of both the points are all positive.!
Due regard being had to the signs of the coordinates, the formula
will be found to be true for all
points.
As a numerical example, let
Pj be the point (5, 6) and Pg
be the point (-7, -4), so that
we have
and y2= -^.
Then
P^ = 31^0 + OM^ = 7 + 5
and
RPt^ = EM-^ + l/iPj = 4 + 6
The rest of the proof is as in the last article.
Similarly any other case could be considered.
22. To find tJie coordinates of the point which divides
in a given ratio (ni^ : m^ the line joining two given jyoints
(a?!, 2/1) and (x^, y^).
Yi
O M, M
M,
X
Let Pi be the point {x^, y^), Po the point (x^, y^), and P
the required point, so that we have
12 COORDINATE GEOMETRY.
Let P be the point (sc, y) so that if P^M^, PM, and
P^M^ be drawn parallel to the axis of y to meet the axis of
£C in i/i, Mj and M^, we have
Oi/i = £Ci, M^P^ = y^, OM=x, MP = y, QM^^x^,
and i/^z^a = 2/2-
Draw PiEi and P-Sg, parallel to OX, to meet J/P and
M^P^ in Pi and Pg respectively.
Then PjPi = M^^M^^ OM- OM^ = x-x^,
PR^ = MM^ = OJ/2 - 0M= x,^ - X,
R,P^MP-M,P, = y-y,,
and P2P2 = M^P^ - MP = y^-y.
. From the similar triangles PiPjP and PR^P^ we have
m^ PjP PiRi X - Xt^
m^ PP^ PR^ x^ - x'
, ifv-tt/Uey *T" i/VoOO-i
Again
mi P,P R,P y-y.
- * m^ PP2 P2P2 2/2-2/'
so that mi (3/2 - 3/) = 7^2 {y - 3/1),
and hence y = -^^ ^-^ .
Wi + 7?22
The coordinates of the point which divides PiP^ in-
ternally in the given ratio rrii : tyi^ are therefore
nil + ^2 mi + nig *
If the point Q divide the line P1P2 externally in the
same ratio, i.e. so that P^Q : QP^ :: mj : m^i its coordinates
would be found to be
The proof of this statement is similar to that of the
preceding article and is left as an exercise for the student.
LINES DIVIDED IN A GIVEN RATIO. 13
Cor. The coordinates of the middle point of the line
joining {x^, y^ to {x^, y^ are
23. Ex. 1. In any triangle ABC 'prove that
AB^ + AC^ = 2 {AD^ + DG^),
lohere D is the middle point of BG.
Take B as origin, 5C as the axis of x, and a line through B i>er-
pendicular to BC as the axis of y.
Let BG=a, so that G is the point (a, 0), and let A be the point
Then D is the point (|> C> j .
Hence ^D2=ra;i -^Y + i/i^ and DG^=f~y.
Hence 2 (^D^ + DC^) ::= 2 ["x^^ + y^^ - ax^ + ^~|
= 2xi2 + 2yi2_2o.x.^ + a2.
Also ^C'2.= (a;i-a)2 + ?j^2^
and AB^=^x^-\-y^.
Therefore AB'^ + ^(72 = 'Ix^ + 2?/i2 _ 2aa;i + a^.
Hence ^52 + ^(72^2(^2)2 + 2)(72)_
This is the well-known theorem of Ptolemy.
Ex. 2. ABG is a triangle and D, E, and F are the middle points
of the sides BG, GA, and AB ; prove that the point lohich divides AD
internally in the ratio 2 : 1 also divides the lines BE and GF in
the same ratio.
Hence prove that the medians of a triangle meet in a point.
Let the coordinates of the vertices A, J5, and G be (x-^, y-^), (arg, 2/2),
and (ajg, y^) respectively.
The coordinates of D are therefore -^ ^ and - ^^ .
Let G be the point that divides internally AD in the ratio 2 : 1,
and let its coordinates be x and y.
By the last article
2Xiy "T Xq
2+1 3
So ^^2/rt^±^3.
14 COORDINATE GEOMETRY.
In the same manner we could shew that these are th^ coordinates
of the points that divide BE and CF in the ratio 2 : 1.
Since the point whose coordinates are
x-^ + x^ + x^ and ^L±^2+l_3
3 3
|
lies on each of the lines AD, BE, and CF, it follows that these three
lines meet in a point.
This point is called the Centroid of the triangle.
EXAMPLES. I.
Find the distances between the following pairs of points.
1. (2, 3) and (5, 7). 2. (4, -7) and (-1, 5).
3_ ( _ 3, _ 2) and ( - 6, 7), the axes being inclined at 60°.
4. (a, o) and (o, 6). 5. {b + c, c + a) and {c + a, a + b).
6. {a cos a, a sin a) and {a cos |S, a sin /3).
7. {am^^, 2ami) and (am^^ 2am^.
8. Lay down in a figure the positions of the points (1, - 3) and
( - 2,1), and prove that the distance between them is 5.
9. Find the value of x^ if the distance between the points [x^^, 2)
and (3, 4) be 8.
10. A line is of length 10 and one end is at the point (2, - 3) ;
if the abscissa of the other end be 10, prove that its ordinate must be
3 or - 9.
11. Prove that the points (2a, 4a), (2a, 6a), and (2a + s/3a, oa)
are the vertices of an equilateral triangle whose side is 2a.
12. Prove that the points (-2, -1), (1, 0), (4, 3), and (1, 2) are
at the vertices of a parallelogram.
13. Prove that the points (2, -2), (8, 4), (5, 7), and (-1, 1) are
at the angular points of a rectangle.
14. Prove that the point ( - xV. f I) is the centre of the circle
circumscribing the triangle whose angular points are (1, 1), (2, 3),
and ( - 2, 2).
Find the coordinates of the point which
15. divides the line joining the points (1, 3) and (2, 7) in the
ratio 3 : 4.
16. divides the same line in the ratio 3 : - 4.
17. divides, internally and externally, the line joining ( - 1, 2)
to (4, - 5) in the ratio 2 : 3.
[EXS. I.] EXAMPLES. 15
18. divides, internally and externally, the line joining ( - 3, - 4)
to ( - 8, 7) in the ratio 7 : 5.
19. The line joining the points (1, - 2) and ( - 3, 4) is trisected ;
find the coordinates of the points of trisection.
20. The line joining the points ( - 6, 8) and (8, - 6) is divided
into four equal parts ; find the coordinates of the points of section.
21. rind the coordinates of the points which divide, internally
and externally, the line joining the point {a + b, a-h) to the point
(a-&, a + 6) in the ratio a : h.
22. The coordinates of the vertices of a triangle are [x-^, 2/i)»
{x^, y^ and (xg, y^. The line joining the first two is divided in the
ratio I : h, and the line joining this point of division to the opposite
angular point is then divided in the ratio m : Jfc + Z. Find the
coordinates of the latter point of section.
23. Prove that the coordinates, x and y, of the middle point of
the line joining the point (2,3) to the point (3, 4) satisfy the equation
x-y + l=:0.
24. If G be the centroid of a triangle ABC and O be any other
point, prove that
^{GA^-^GB'^+GCr-) = BG^+GA^ + AB\
and OA^ + OB^-\-OG'^=GA^ + GB'^+GG^- + ^Ga\
25. Prove that the lines joining the middle points of opposite
sides of a quadrilateral and the Une joining the middle points of its
diagonals meet in a point and bisect one another.
26. -4, B, G, D... are n points in a plane whose coordinates are
(^i» 2/i)> (^2' 2/2)' (^3> 2/3)j---* -^-S is bisected in the point G-^; G^G is
divided at G^ in the ratio 1:2; G^D is divided at G^ in the ratio
1:3; GgE at G^ in the ratio 1 : 4, and so on until all the points are
exhausted. Shew that the coordinates of the final point so obtained are
n n
[This point is called the Centre of Mean Position of the n given
points.]
27. Prove that a point can be found which is at the same
distance from each of the four points
(am,, ^) , (a»„ ^) , {am,, ^J , and {am^m^, ^^) .
24. To prove that the area of a trapeziitm, i. e. a quad-
rilateral having two sides parallel, is one half the sum of the
two parallel sides multiplied by the perpendicular distance
between them.
COOEDINATE GEOMETRY.
B
Let ABGD be the trapezium having the sides AD and
BC parallel.
Join AC and draw AL perpen-
dicular to BG and ON perpendicular
to AD^ produced if necessary.
|
Since the area of a triangle is one
half the product of any side and the
perpendicular drawn from the opposite angle, we have
area ABGD = ^ABG ■¥ ^AGD
= l.BG ,AL + l,AD.GN
=^i{BG + AD) X AL.
25. To find the area of the triangle^ the coordinates of
whose angular 'points are given^ the axes being rectangular.
Let ABG be the triangle
and let the coordinates of its
angular points A, B and G be
{x^, 2/1), (a?2, 2/2), and {x^, y^).
Draw AL, BM, and Ci\^ per-
pendicular to the axis of x, and
let A denote the required area.
Then
A == trapezium A LNG + trapezium GNMB - trapezium A LMB
= \LN {LA + NG) + \NM {NG + MB) - \LM (LA + MB),
by the last article,
= i [(^3 - ^1) (2/1 + ys) + {002 - ^z) (2/2 + 2/3) - (^2 - a^i) {Vi + 2/2)]-
On simplifying we easily have
^ ^ I (^172 - XaYi + y^zYz - ^372 + XgYi - x^yg),
or the equivalent form
^ = J [^1 (2/2 - Vz) + ^2 (2/3 - 2/1) + ^3 (2/1 - 2/2)].
If we use the determinant notation this may be written
(as in Art. 5)
, ^3j 2/35 ^
Cor. The area of the triangle whose vertices are the
origin (0, 0) and the points {x^, y-^, {x^, 2/2) is J (^12/2 - ^'22/1)*
AREA OF A QUADRILATERAL,
isin w {x^y^
I.e.
fsm wx
26. In the preceding article, if the axes be oblique, the perpen-
diculars AL, BM, and CN, are not equal to the ordinates y^ , y^ , and
2/3, but are equal respectively to yi sin w, 2/2 sin w, and y^ sin w.
The area of the triangle in this case becomes
xu yi, 1
27. In order that the expression for the area in Art. 25 may be
a positive quantity (as all areas necessarily are) the points A, B, and
G must be taken in the order in which they would be met by a
person starting from A and walking round the triangle in such a
manner that the area of the triangle is always on his left hand.
Otherwise the expressions of Art. 25 would be found to be negative.
28. To find the area of a quadrilateral the coordinates
of whose angular foints are given.
Let the angular points of the quadrilateral, taken in
order, be A^ B, C, and D, and let their coordinates be
respectively {x^, y^\ (x.^,, y^), (x^, y^\ and {x^, y^).
Draw ALy BM, CJV, and DH perpendicular to the axis
of X.
Then the area of the quadrilateral
= trapezium ALRD + trapezium BRNO + trapezium GNMB
- trapezium ALMB
= ILR {LA + RD) + IRN (RD + NG) + ^NM {JVC + MB)
-lLM{LA-\-MB)
{(^■4 - ^1) (2/1 + 2/4) + (^3 - ^4) (2/3 + 2/4) + (^2 - ^^) (2/3 + 2/2)
{(^12/2 - «^22/i) + fez^s - ^zvi) + {xsy4. - ^42/3) + (^42/i - ^y^)}'
L. 2
18 COORDINATE GEOMETRY.
29. The above formula may also be obtained by
drawing the lines OA, OB, OC and OD. For the quadri-
lateral ABCn
= AOBG+ AOCD- aOBA- AOAD.
But the coordinates of the vertices of the triangle OBG
are (0, 0), (ajg, 2/2) ^^^ (^35 2/3) ^ hence, by Art. 25, its
area is ^ i^^y-^ - ^zV^)-
So for the other triangles.
The required area therefore
^ h [(«^22/3 - a^32/2) + {^zVa, - ^m) - (^Wi - ^12/2) - {^i2/4 - ^'42/l)l
= i [{^^2 - ^22/1) + (^22/3 - ^32/2) + {^3y4 - ^m) + (^42/l - «^l2/4)]-
In a similar manner it may be shewn that the area
of a polygon of n sides the coordinates of whose angular
points, taken in order, are
(^IJ 2/1/5 V^2) 2/2/3 (^35 2/3)? •••(.'^)i3 2/ra/
is i [(a?i2/2 - a?22/i) + (^22/3 - ^32/2) + • • • + (^n2/i - a'l^/ri)]-
EXAMPLES. II.
Find the areas of the triangles the coordinates of whose angular
points are respectively
1. (1, 3), ( - 7, 6) and (5, - 1). 2. (0, 4), (3, 6) and ( - 8, - 2).
3. (5,2), (-9, -3) and (-3, -5).
4. {a, l> + c), (a, h-c) and {-a, c).
5. {a,c + a), {a, c) and {-a, c-a).
6. {a cos (pi, b sin <p-^), {a cos ^^, b sin ^g) and (a cos ^3, b sin ^g).
7. (a7?ij2^ 2a7?i;^), [arn^^ lam^ and [am^, 2avi^.
8. {awiimg, a(??ii + ??i2)}, {aWaWg, a (7?i2 + %)} ^-^cl
9. lam-,. - I , \am^, - y and -Jawo, - [ .
Prove (by shewing that the area of the triangle formed by them is
zero) that the following sets of three points are in a straight line :
10. (1,4), (3, -2), and (-3,16).
11. (-i, 3), (-5,6), and (-8,8).
[EXS. II.] POLAR COORDINATES. 19
|
Find the areas of the quadrilaterals the coordinates of whose
angular points, taken in order, are
13. (1,1), (3,4), (5, -2), and (4, -7).
15. If 0 be the origin, and if the coordinates of any two points
Pj and Pg ^6 respectively (%, y^ and [x^, y.^, prove that
OP^ . OP2 . cos P1OP2 = x-^Xc^ + y-^y^ •
30. Polar Coordinates. There is another method,
which is often used, for determining the position of a point
in a plane.
Suppose 0 to be a fixed point, called the origin or
pole, and OX a fixed line, called the initial line.
Take any other point P in the plane of the paper and
join OP. The position of P is clearly known when the
angle XOP and the length OP are given.
[For giving the angle XOP shews the direction in which OP is
drawn, and giving the distance OP tells the distance of P along this
direction.]
The angle XOP which would be traced out by the line
OP in revolving from the initial line OX is called the
vectorial angle of P and the length OP is called its radius
vector. The two taken together are called the polar co-
ordinates of P.
If the vectorial angle be 0 and the radius vector be r, the
position of P is denoted by the symbol (r, 0).
The radius vector is positive if it be measured from the
origin 0 along the line bounding the vectorial angle; if
measured in the opposite direction it is negative.
31. Ex. Construct the positions of the 'points (i) (2, 80°),
(ii) (3, 150°), (iii) (-2, 45°), (iv) ■
(-3, 330°), (v) (3, -210°) and (vi) ff\ /^
(i) To construct the first point, ^^^^^^^^^"^^
let the radius vector revolve from '/^^^ yC
OX through an angle of 30°, and >/ ""-.,
then mark off along it a distance y ''-•.,
equal to two units of length. We 'p M"
thus obtain the point P^. ^
(ii) For the second point, the radius vector revolves from OX
through 150° and is then in the position OP^ ; measuring a distance 3
along it we arrive at Pg .
2 - 2
20 COORDINATE GEOMETKY.
(iii) For the third point, let the radius vector revolve from OX
through 45° into the position OL. We have now to measure along
OL a distance - 2, i.e. we have to measure a distance 2 not along OL
but in the opposite direction. Producing iO to Pg, so that OP3 is
2 units of length, we have the required point P3.
(iv) To get the fourth point, we let the radius vector rotate from
OX through 330° into the position OM and measure on it a distance
-3, i.e. 3 in the direction MO produced. We thus have the point P^y
which is the same as the point given by (ii).
(v) If the radius vector rotate through - 210°, it will be in the
position OP2, and the point required is Pg.
(vi)^ For the sixth point, the radius vector, after rotating through
- 30°, is in the position OM: We then measure - 3 along it, i.e. 3 in
the direction MO produced, and once more arrive at the point Pg.
32. It will be observed that in the previous example
the same point P^ is denoted by each of the four sets of
polar coordinates
(3, 150°), (-3, 330°), (3, -210°) and (-3, -30°).
In general it v^ill be found that the same point is given
by each of the polar coordinates
(r, 0), (- r, 180° + 6), {r, - (360° - 6)] and {- r, - (180° - 6%
or, expressing the angles in radians, by each of the co-
ordinates
(r, e\ {-r,7r + 6), {r, - (27r - 0)} and {- r, - (tt - $)}.
It is also clear that adding 360° (or any multiple of
360°) to the vectorial angle does not alter the final position
of the revolving line, so that {r, 6) is always the same point
as (r, ^ + ?i . 360°), where n is an integer.
So, adding 180° or any odd multiple of 180° to the
vectorial angle and changing the sign of the radius vector
gives the same point as before. Thus the point
[-r, ^ + (2n + 1)180°]
is the same point as [ - r, 6 + 180°], i.e. is the point [r, 6\
33. To find the length of the straight line joining two
points whose polar coordinates are given.
Let A and B be the two points and let their polar
coordinates be (r^, 6y) and (r^, 6^ respectively, so that
OA^r^, OB = r^, lXOA^O^, and lX0B = 6^,
|
POLAR COORDINATES.
Then (Trigonometry, Art. 164)
AB" - OA'' + OB'' -20 A. OB cos AOB
34. To find the area of a triangle the coordinates of
whose angular points are given.
Let ABC be the triangle and let (r-^, 0^), (r^, 62), and
(rg, ^3) be the polar coordinates of
its angular points.
We have
AABO=AOBC+aOCA
-AOBA (1).
Now
A0BC = i0B,0C sin BOC
[Trigonometry, Art. 198]
' So A OCA = \0G . OA sin CO A = ^r^r, sin (6, - 6,),
and AOAB^^OA. OB sin AOB = ^r^r^ sin {6^ - 6.^
= - Jn^2 sin (^2 - ^1).
Hence (1) gives
A ABC = J \r<^r^ sin (^3 - 6^ + r^r^ (sin 0-^ - 0^)
+ r^r^ sin {Oo - 0^)].
35. To change from Cartesian Coordinates to Polar
Coordinates, and conversely.
Let P be any point whose Cartesian coordinates, referred
to rectangular axes, are x and y,
and whose polar coordinates, re-
ferred to 0 as pole and OX as
initial line, are (r, 6).
Draw Pit/'perpendicular to OX
so that we have
OM=x, MP = y, LMOP = e,
and OP = r.
From the triangle MOP we
have
x = OM=OPcosMOP = rcosO (1),
y = MP=OPsinMOP^rsmO (2),
r=OP= sJOM^ + MP^^ s/x" + y' (3),
X' O:
22 COORDINATE GEOMETRY,
and
Equations (1) and (2) express the Cartesian coordinates
in terms of the polar coordinates.
Equations (3) and (4) express the polar in terms of the
Cartesian coordinates.
The same relations will be found to hold if P be in any-
other of the quadrants into which the plane is divided by
XOX' and YOT.
Ex. Change to Cartesian coordinates the equations
{!) r = asind, and (2)r=a^cos-.
a
(1) Multiplying the equation by r, it becomes r^rrar sin Q,
i.e. by equations (2) and (3), x^-\-y^ = ay.
(2) Squaring the equation (2), it becomes
r=acos2- = -- (1 + cos^),
i. e. 2^2 = ar + ar cos 6,
i.e. 2{x^ + y^) = a sjx^ + y^-\- ax,
i.e. {2x^ + 2y^-ax)^ = a^{x^ + y^).
EXAMPLES. III.
Lay down the positions of the points whose polar coordinates are
L (3,45°). 2. (-2, -60°). 3. (4,135°). 4. (2,330°).
5. (-1, -180°). 6. (1, -210°). 7. (5, -675°). 8. («> |) •
9. (2a.-^). 10.{-a,l). n.(-2a.4').
Find the lengths of the straight lines joining the pairs of points
whose polar coordinates are
12. (2, 30°) and (4, 120°). 13. (-3, 45°) and (7, 105°).
14. ( «> I ) and
[EXS. III.] EXAMPLES. 23
15. Prove that the points (0, 0), (3, |) , and ( 3, ^ J form an equi-
lateral triangle.
Find the areas of the triangles the coordinates of whose angular
points are
16. (1, 30°), (2, 60°), and (3, 90°).
17. (_3, -30°), (5, 150°), and (7, 210°).
Find the polar coordinates (drawing the figure in each case) of the
points
19. x = J3, y = l. 20. x=-^B, y = l. 21. x=-l, y = l.
Find the Cartesian coordinates (drawing a figure in each case) of
the points whose polar coordinates are
22. (5, I). 23. (-6, I). 24.(5,-^).
Change to polar coordinates the equations
25. x^+y^=a\ 26. y = xtana. 27. x^ + y^ = 2ax.
28. x^-y^=2ay. 29. x^=y^{2a-x). 30. {x^ + y^)^ = a^{x^-y^).
Transform to Cartesian coordinates the equations
31. r=a. 32. ^ = tan-i??i. 33. r = acos^.
34. r = asin2^. 35. ?-2 = a2cos2^. 36. o'^ sin 2d =2a^.
e
40. 'T (cos 3^ + sin 3^) = 5h sin 6 cos d
37. r'^G0s2d = a^. 38. r^cos- = a^, 39^ r^=ai sin-.
2i a
CHAPTER III.
LOCUS. EQUATION TO A LOCUS.
36. When a point moves so as always to satisfy a
given condition, or conditions, the path it traces out is
called its Locus under these conditions.
For example, suppose 0 to be a given point in the plane
of the paper and that a point F is to move on the paper so
that its distance from 0 shall be constant and equal to a.
It is clear that all the positions of the moving point must
lie on the circumference of a circle whose centre is 0 and
whose radius is a. The circumference of this circle is
therefore the " Locus" of P when it moves subject to the
condition that its distance from 0 shall be equal to the
constant distance a.
|
37. Again, suppose A and B to be two fixed points in
the plane of the paper and that a point P is to move in
the plane of the paper so that its distances from A and B
are to be always equal. If we bisect AB in G and through
it draw a straight line (of infinite length in both directions)
perpendicular to AB, then any point on this straight line
is at equal distances from A and B. Also there is no
point, whose distances from A and B are the same, which
does not lie on this straight line. This straight line is
therefore the "Locus" of P subject to the assumed con-
dition.
38. Again, suppose A and B to be two fixed points
and that the point P is to move in the plane of the paper
so that the angle APB is always a right angle. If we
describe a circle on AB as diameter then P may be any
EQUATION TO A LOCUS.
point on the circumference of this circle, since the angle
in a semi-circle is a right angle; also it could easily be
shewn that APB is not a right angle except when P lies
on this circumference. The "Locus" of P under the
assumed condition is therefore a circle on -4^ as diameter.
39. One single equation between two unknown quan-
tities x and y, e.g.
cannot completely determine the values of x and y.
*\P6
\Q
vPs
\1
M
OM
Vs
Such an equation has an infinite number of solutions.
Amongst them are the followins: :
a: = 0,1
a; =1,1
x= 2,1
x= 3,1
2/= 2/' y= 3
Let us mark down on paper a number of points whose
coordinates (as defined in the last chapter) satisfy equation
Let OX and OF be the axes of coordinates.
If we mark off a distance OPi ( - 1) along OY^ we have
a point Pi whose coordinates (0, 1) clearly satisfy equation
If we mark off a distance OP^ (=1) along OX, we have
a point Pg whose coordinates (1, 0) satisfy (1).
26 COORDINATE GEOMETRY.
Similarly the point Pg, (2, - 1), and P4, (3, - 2), satisfy
the equation (1).
Again, the coordinates ( - 1, 2) of Pg and the coordinates
( - 2, 3) of Pg satisfy equation (1).
On making the measurements carefully we should find
that all the points we obtain lie on the line P^P^ (produced
both ways).
Again, if we took any point Q, lying on P^P^, and draw
a perpendicular QM to OX, we should find on measurement
that the sum of its x and y (each taken with its proper
sign) would be equal to unity, so that the coordinates of Q
would satisfy (1).
Also we should find no point, whose coordinates satisfy
(1), which does not lie on P1P2.
All the points, lying on the straight line P1P2, and no
others are therefore such that their coordinates satisfy the
equation (1).
This result is expressed in the language of Analytical
Geometry by saying that (1) is the Equation to the Straight
Line P^P^.
40. Consider again the equation
£c2 + 2/2 = 4 (1).
Amongst an infinite number of solutions of this equa-
tion are the following :
x = 2A x= J?>\ x = J\ x^l \
V3| x = J'I\ x=^i \
2/=2r 2/=x/3r 2/=V2 r y=i r
2/^0 r 2/^--i /' 3/=-v2r y=-si^)'
53 = 0, \ x=l, \ x=J2, } x=J3)
2/ = -2j ' y
L and
EQUATION TO A LOCUS.
,5i
■i|f
0 M ;f^ X
All these points are respectively represented by the
points P^, P^^ F^y ... P^Qj and they
will all be found to lie on the
dotted circle whose centre is 0
and radius is 2.
Also, if we take any other
point Q on this circle and its
ordinate QM, it follows, since
0M^- + MQ^^0Q' = 4:, that the x
and y of the point Q satisfies (1).
The dotted circle therefore
passes through all the points whose
coordinates satisfy (1).
In the language of Analytical Geometry the equation
(1) is therefore the equation to the above circle.
41. As another example let us trace the locus of the
point whose coordinates satisfy the equation
If we give x a negative value we see that y is im-
possible; for the square of a
real quantity cannot be nega-
tive.
We see therefore that there
are no points lying to the left
of OF.
If we give x any positive
value we see that y has two
real corresponding values which
are equal and of opposite signs.
|
The following values,
amongst an infinite number of
others, satisfy (1), viz.
x = 0,] x=l, "I x = 2,
y = 0}' y = + 2or-2}' y = 2J2ov -2J2
x=4: I cc=16, I aj = + oo, )
y = + 4: or - 4:) ^ '" y - Sor-Sj' '" y = + (X)Ov - ao)'
The origin is the first of these points and P^ and Qi,
Pg and Q^, P^and Q^, ... represent the next pairs of points.
h
28 COORDINATE GEOMETRY.
If we took a large number of values of x and the
corresponding values of ?/, the points thus obtained would
be found all to lie on the curve in the figure.
Both of its branches would be found to stretch away to
infinity towards the right of the figure.
Also, if we took any point on this curve and measured
with sufficient accuracy its x and y the values thus obtained
would be found to satisfy equation (1).
Also we should not be able to find any point, not lying
on the curve, whose coordinates would satisfy (1).
In the language of Analytical Geometry the equation
(1) is the equation to the above curve. This curve is called
a Parabola and will be fully discussed in Chapter X.
42. If a point move so as to satisfy any given condition
it will describe some definite curve, or locus, and there can
always be found an equation between the x and y of any
point on the path.
This equation is called the equation to the locus or
curve. Hence
Def. Equation to a curve. The equation to a
curve is the relation which exists between the coordinates of
any foint on the curve^ and which holds for no other points
except those lying on the curve.
43. Conversely to every equation between x and y it
will be found that there is, in general, a definite geometrical
locus.
Thus in Art. 39 the equation is x + y=\, and the
definite path, or locus, is the straight line P^P^ (produced
indefinitely both ways).
In Art. 40 the equation is x'^ + y'^^ 4, and the definite
path, or locus, is the dotted circle.
Again the equation 2/ = 1 states that the moving point
is such that its ordinate is always unity, i.e. that it is
always at a distance 1 from the axis of x. The definite
path, or locus, is therefore a straight line parallel to OX
and at a distance unity from it.
EQUATION TO A LOCUS. 29
44. In the next chapter it will be found that if the
equation be of the first degree {i.e. if it contain no
products, squares, or higher powers of x and y) the locus
corresponding is always a straight line.
If the equation be of the second or higher degree, the
corresponding locus is, in general, a curved line.
45. We append a few simple examples of the forma-
tion of the equation to a locus.
Ex. 1. A point moves so that the algebraic sum of its distances
from tioo given perpendicular axes is equal to a constant quantity a;
find the equation to its locus.
Take the two straight lines as the axes of coordinates. Let {x, y)
be any point satisfying the given condition. We then ha,wex + y = a.
This being the relation connecting the coordinates of any point
on the locus is the equation to the locus.
It will be found in the next chapter that this equation represents
a straight line.
Ex. 2. The sum of the squares of the distances of a moving point
from the tioo fixed points {a, 0) and {-a, 0) is equal to a constant
quantity 2c^. Find the equation to its locus.
Let (a;, y) be any position of the moving point. Then, by Art. 20,
the condition of the question gives
This being the relation between the coordinates of any, and every,
point that satisfies the given condition is, by Art. 42, the equation to
the required locus.
This equation tells us that the square of the distance of the point
{x, y) from the origin is constant and equal to c^ - a^, and therefore
the locus of the point is a circle whose centre is the origin.
Ex. 3. A point moves so that its distance from the point (-1,0)
is always three times its distance from the point (0, 2).
Let {x, y) be any point which satisfies the given condition. We
then have
J{x + iy' + {y-0)^=Bj{x - 0)2+ {y - 2)2,
so that, on squaring,
|
i.e. 8(a;2 + y2)_2a;-36?/ + 35 = 0.
This being the relation between the coordinates of each, and
every, point that satisfies the given relation is, by Art. 42, the
required equation.
It will be found, in a later chapter, that this equation represents
a circle.
30 COOEDINATE GEOMETRY.
EXAMPLES. IV.
By taking a number of solutions, as in Arts. 39 - 41, sketch
the loci of the following equations :
1. 2x + dy = l0. 2. ^x-y = l. 3. x'^-2ax-Vy'^ = Q.
4. a;2-4aa; + ?/2 + 3a2 = 0. 5. y'^ = x. 6. ^x = y^-^.
'■4^9
A and B being the fixed points (a, 0) and ( - a, 0) respectively,
obtain the equations giving the locus of P, when
8. PA"^ - P52 _ a constant quantity = 2fc2.
9. PA = nPB, n being constant.
10. P^+PjB = c, a constant quantity.
11. PB^ + PC^=2PA^, C being the point (c, 0).
12. Find the locus of a point whose distance from the point (1, 2)
is equal to its distance from the axis of y.
Find the equation to the locus of a point which is always equi-
distant from the points whose coordinates are
13. (1, 0) and (0, -2). 14. (2, 3) and (4, 5).
15. {a + b, a-h) and {a-b, a + b).
Find the equation to the locus of a point which moves so that
16. its distance from the axis of x is three times its distance from
the axis of y.
17. its distance from the point (a, 0) is always four times its dis-
tance from the axis of y.
18. the sum of the squares of its distances from the axes is equal
to 3.
19. the square of its distance from the point (0, 2) is equal to 4.
20. its distance from the point (3, 0) is three times its distance
from (0, 2).
21. its distance from the axis of x is always one half its distance
from the origin.
22. A fixed point is at a perpendicular distance a from a fixed
straight line and a point moves so that its distance from the fixed
point is always equal to its distance from the fixed line. Find the
equation to its locus, the axes of coordinates being drawn through
the fixed point and being parallel and perpendicular to the given
line.
23. In the previous question if the first distance be (1), always half,
and (2), always twice, the second distance, find the equations to the
respective loci.
CHAPTER IV.
THE STRAIGHT LINE. RECTANGULAR COORDINATES.
46. To find the equation to a straight line which is
parallel to one of the coordinate axes.
Let CL be any line parallel to the axis of y and passing
through a point C on the axis of x such that OG = c.
Let F be any point on this line whose coordinates are
X and y.
Then the abscissa of the point F is
always c, so that
This being true for every point on
the line CL (produced indefinitely both
ways), and for no other point, is, by
Art. 42, the equation to the line.
It will be noted that the equation does not contain the
coordinate y.
Similarly the equation to a straight line parallel to the
axis oi X is y - d.
Cor. The equation to the axis of a? is 2/ = 0.
The equation to the axis oi y is x - 0.
47. To find the equation to a st7'aight line which cuts
off a given intercept on the axis of y and is inclined at a
given angle to the axis of x.
Let the given intercept be c and let the given angle be a.
X
32 COORDINATE GEOMETRY.
Let C be a point on the axis of y such that OC is c.
Through C draw a straight
line Z(7Z' inclined at an angle
a (= tan~^ m) to the axis of x^
so that tan a - m.
The straight line LCL' is ^^
therefore the straight line ^^
required, and we have to -'l O MX
find the relation between the
coordinates of any point P lying on it.
Draw PM perpendicular to OX to meet in ^ a line
through G parallel to OX.
Let the coordinates of P be cu and ?/» so that OM=x
and MP = y.
Then MP = NP + MN =C]Srt^iia + 00 = m.x + c,
i.e. y = mx+c.
This relation being true for any point on the given
straight line is, by Art. 42, the equation to the straight
line.
[In this, and other similar cases, it could be shewn,
conversely, that the equation is only true for points lying
on the given straight line.]
|
Cor. The equation to any straight line passing through
the origin, i.e. which cuts off a zero intercept from the axis
of 2/, is found by putting c - O and hence is 3/ = mx.
48. The angle a which is used in the previous article is the
angle through which a straight line, originally parallel to OZ, would
have to turn in order to coincide with the given direction, the rotation
being always in the positive direction. Also m is always the tangent
of this angle. In the case of such a straight line as AB, in the figure
of Art. 50, m is equal to the tangent of the angle PAX (not of the
angle PAO). In this case therefore wi, being the tangent of an obtuse
angle, is a negative quantity.
The student should verify the truth of the equation of the last
article for all points on the straight line LCL', and also for straight
Hnes in other positions, e.g. for such a straight line as A^B^ in the
figure of Art. 59. In this latter case both m and c are negative
quantities.
A careful consideration of all the possible cases of a few proposi-
tions will soon satisfy him that this verification is not always
necessary, but that it is sufficient to consider the standard figure.
THE STRAIGHT LINE.
49. Ex. The equation to the straight line cutting off an
intercept 3 from the negative direction of the axis of y, and inclined
at 120° to the axis of a;, is
?/ = a;tanl20° + (-3),
i.e. y= -x^S-S,
i.e. y + x^S + S = 0.
50. 1^0 find the equation to the straight line which cuts
off given i7itercepts a and h from the axes.
Let A and B be on OX and OY respectively, and be
such that OA = a and OB = h.
Join AB and produce it in-
definitely both ways. Let P be
any point (x, y) on this straight
line, and draw PM perpendicular
to OX.
We require the relation that
always holds between x and 3/, so
long as P lies on AB.
By Euc. YI. 4, we have
OM_PB MP _AP
OA~AB' ^"""^ 'OB~AB
OM MP PB + AP
OA OB
AB
X y ^
a D
This is therefore the required equation ; for it is the
relation that holds between the coordinates of any point
lying on the given straight line.
51. The equation in the preceding article may he also obtained
by expressing the fact that the sum of the areas of the triangles OP A
and OPB is equal to OAB, so that
\axy + \hy.x = \ax'b,
and hence
a 0
52. Ex. 1. Find the equation to the straight line passing
through the -point (3, - 4) and cutting off intercepts, equal but of
opposite signs, from the tioo axes.
Let the intercepts cut off from the two axes be of lengths a and
COORDINATE GEOMETRY.
The equation to the straight line is then
a -a
i.e. x-y = a (1).
Since, in addition, the straight line is to go through the point
(3, -4), these coordinates must satisfy (1), so that
and therefore a = l.
The required equation is therefore
x-y = 7.
Ex. 2. Find the equation to the straight line lohich passes through
the point (-5, 4) and is such that the portion of it between the axes is
divided by the point in the ratio ofl : 2.
Let the required straight line be - + t = 1. This meets the axes
a b
in the points whose coordinates are {a, 0) and (0, &).
The coordinates of the point dividing the line joining these
points in the ratio 1 : 2, are (Art. 22)
2.a+1.0 , 2.0 + 1.&
If this be the point ( - 5, 4) we have
. 2(1 , b
i.e,-^ and -.
-5:=- and 4=-,
so that a=--Y- and b = 12.
The required straight line is therefore
X y
-i^^l2 '
I.e.
oy
8a; = 60.
53. To find the equation to a straight line in tenns of
the perpendicular let fall upon it from the origin and the
angle that this perpendicular makes with the axis of x.
Let OR be the perpendicular from 0 and let its length
be jo.
Let a be the angle that OR makes
with OX.
Let P be any point, whose co-
ordinates are x and y, lying on AB ',
draw the ordinate PM, and also ML
perpendicular to OR and PN perpen-
dicular to ML.
THE STRAIGHT LINE. 85
Then OL = OMco^a (1),
and LR = NP = MF&inNMP.
But lNMP^W - lNMO= iMOL^a.
LR = MP&m.a (2).
Hence, adding (1) and (2), we have
|
Oil/ cos a + ifPsin a=OL + LR=OR =79,
i.e. X COS a + y sin a = p.
This is the required equation.
54. In Arts. 47 - 53 we have found that the correspond-
ing equations are only of the first degree in x and y. We
shall now prove that
Any equation of the first degree i7i x and y always repre-
sents a straight line.
For the most general form of such an equation is
Ax + By^C = ^ (1),
where A^ B, and C are constants, i.e. quantities which do
not contain x and y and which remain the same for all
points on the locus.
Let (cCi, 2/1), (a?2) 2/2)) ^iicl (rt's, 2/3) be any three points on
the locus of the equation (1).
Since the point {x-^, y^) lies on the locus, its coordinates
when substituted for x and y in (1) must satisfy it.
Hence Ax^ + Ry^+C-=0 (2).
So Ax^ + Ry^ + C^O (3),
and Axs + £ys+C = 0 (4).
Since these three equations hold between the three quanti-
ties A, B, and C, we can, as in Art. 12, eliminate them.
The result is
^35 2/35 -•-
But, by Art. 25, the relation (5) states that the area of the
triangle whose vertices are (x^, y^), (x^, 3/2)5 ^^^ (^3> 2/3) is
zero.
Also these are any three points on the locus.
3 - 2
36 COORDINATE GEOMETRY.
The locus must therefore be a straight line ; for a curved
line could not be such that the triangle obtained by joining
any three points on it should be zero.
55. The proposition of the preceding article may also be deduced
from Art. 47. For the equation
A C
may be written y=- - x-^,
and this is the same as the straight line
y = mx + c,
A ^ C
if ?3i=- - and c = - - .
x> is
But in Art. 47 it was shewn that y = mx + c was the equation to
a straight line cutting off an intercept c from the axis of y and
inclined at an angle tan~^m to the axis of x.
The equation Ax + By + C=0
C
therefore represents a straight line cutting off an intercept - - from
x>
the axis of y and inclined at an angle tan~^ ( - - | to the axis of x.
56. We can reduce the general equation of the first
degree Ax + By + C = 0 (1)
to the form of Art. 53.
For, if p be the perpendicular from the origin on (1)
and a the angle it makes with the axis, the equation to the
straight line must be
X cos a 4- 2/ sin a - /» = 0 (2).
This equation must therefore be the same as (1).
cos a sin a - p
Hence
ABC
p cos a sin a \/cos^ a + sin^ a 1
C -A -B Ja^ + B' sJa^' + B^
Hence
cos a - - , sm a = , , and p =
s/A^ + B^' \fA-' + B'' sfA^ + B^
The equation (1) may therefore be reduced to the form (2)
by dividing it by JA^ + B^ and arranging it so that the
constant term is negative.
THE STRAIGHT LINE. 37
57. Ex. Reduce to tlie perpendicular form the equation
Here JA'' + B^= ^TTs = sJ4:=2.
Dividing (1) by 2, we have
i.e. X cos 240° + y sin 240° - 1 = 0.
58. To trace the straight line given hy an equation of
the first degree.
Let the equation be
Ax + By + G = 0 (1).
(a) This can be written in the form
A B
Comparing this with the result of Art. 50, we see that it
(J
represents a straight Hne which cuts off intercepts - -^ and
- - from the axes. Its position is therefore known.
jO
If G be zero, the equation (1) reduces to the form
A
and thus (by Art. 47, Cor.) represents a straight hne
passing through the origin inchned at an angle tan~^ I ~ r )
to the axis of x. Its position is therefore known.
(^) The straight line may also be traced by firnding
the coordinates of any two points on it.
G
If we put y - 0 in (1) we have x - - -r. The point
JL
therefore lies on it.
COORDINATE GEOMETRY.
G
If we put oj = 0, we have 2/ = - ^ , so that the point
G^
lies on it.
line.
Hence, as before, we have the position of the straight
69. Ex. Trace the straight lines
(1) 3a;-4i/ + 7 = 0; (2) 7a; + 8y + 9 = 0j
(1) Putting 2/ = 0, we have rc= -|,
and putting x = Q, we have y = ^.
Measuring 0A-^{= -^) along the axis of x we have one point on
the Hne.
Measuring OB^ (=t) along the axis of y we have another point.
Hence A-^B^ , produced both ways, is the required line,
|
(2) Putting in succession y and x equal to zero, we have the
intercepts on the axes equal to - f and - f.
If then 0-42= -f and 0^2= - |, we have A^B^, the required line.
(3) The point (0, 0) satisfies the equation so that the origin is on
the line.
Also the point (3, 1), i.e. C.^, lies on it. The required line is
therefore OC3.
(4) The line ic = 2 is, by Art. 46, parallel to the axis of y and passes
through the point A^ on the axis of x such that 0A^ = 2.
(5) The line y= - 2 is parallel to the axis of x and passes through
the point B^ on the axis of y, such that 0B^= - 2.
60. Straight Line at Infinity. We have seen
that the equation Ax + By + (7 = 0 represents a straight line
STRAIGHT LINE JOINING TWO POINTS. 39
c c
which cuts oiF intercepts - - and - - from the axes of
Ji. Jj
coordinates.
If A vanish, but not B or C, the intercept on the axis
of X is infinitely great. The equation of the straight line
then reduces to the form y = constant, and hence, as in
Art. 46, represents a straight line parallel to Ox.
So if B vanish, but not A or C, the straight line meets
the axis of y at an infinite distance and is therefore parallel
to it.
If A and B both vanish, but not C, these two in-
tercepts are both infinite and therefore the straight line
Q .x + 0 .y + C = 0 is altogether at infinity.
61. The multiplication of an equation by a constant
does not alter it. Thus the equations
2a;-32/+5 = 0 and 10a;- 152/+ 25 - 0
represent the same straight line.
Conversely, if two equations of the first degree repre-
sent the same straight line, one equation must be equal to
the other multiplied by a constant quantity, so that the
ratios of the corresponding coefficients must be the same.
For example, if the equations
a^x + \y + Ci = 0 and A-^^x + B^y + Cj = 0
we must have
«! \ CjL
62. To jind the equation to the straight line which
passes through the two given points {x\ y') and (x", y").
By Art. 47, the equation to any straight line is
y-- mx -VG (1).
By properly determining the quantities m and c we can
make (1) represent any straight line we please.
If (1) pass through the point (a;', y')^ we have
2/' = mas' + c (2).
Substituting for c from (2), the equation (1) becomes
y-y' = m(x-x') (3).
40 COOKDINATE GEOMETRY.
This is the equation to the line going through (x\ y') making
an angle tan~^ m with OX. If in addition (3) passes through
the point {x", y"), then
y - y=m{x - x),
ti r
y -y
Substituting this value in (3), we get as the required
equation
63. Ex. Find the equation to the straight line which passes
through the points (-1, 3) and (4, -2).
Let the required equation be
y=mx + c (1).
Since (1) goes through the first point, we have
3=-m + c, so that c = m + S.
Hence (1) becomes
y = mx + m + S (2).
If in addition the line goes through the second point, we have
-2 = 47?i + m + 3, so that m= -1.
Hence (2) becomes
y=-x + 2, i.e. x + y = 2.
Or, again, using the result of the last article the equation is
i.e. y + x-=2.
64. To fix definitely the position of a straight line we
must have always two quantities given. Thus one point
on the straight line and the direction of the straight line
will determine it; or again two points lying on the straight
line will determine it.
Analytically, the general equation to a straight line
will contain two arbitrary constants, which will have to be
determined so that the general equation may represent any
particular straight line.
Thus, in Art. 47, the quantities m and c which remain
the same, so long as we are considering the same straigld
line, are the two constants for the straight line.
EXAMPLES. 41
Similarly, in Art. 50, the quantities a and h are the
constants for the straight line.
65. In any equation to a locus the quantities x and y,
which are the coordinates of any point on the locus, are
called Current Coordinates ; the curve may be conceived as
traced out by a point which " runs " along the locus.
EXAMPLES. V.
Find the equation to the straight line
|
1. cutting off an intercept unity from the positive direction of the
axis of y and inclined at 45° to the axis of x.
2. cutting off an intercept - 5 from the axis of y and being equally
inclined to the axes.
3. cutting off an intercept 2 from the negative direction of the
axis of y and inclined at 30° to OX.
4. cutting off an intercept - 3 from the axis of y and inclined at
an angle tan~i f to the axis of x.
Find the equation to the straight line
5. cutting off intercepts 3 and 2 from the axes.
6. cutting off intercepts - 5 and 6 from the axes.
7. Find the equation to the straight line which passes through the
point (5, 6) and has intercepts on the axes
(1) equal in magnitude and both positive,
(2) equal in magnitude but opposite in sign.
8. Find the equations to the straight lines which pass through
the point (1, - 2) and cut off equal distances from the two axes.
9. Find the equation to the straight line which passes through
the given point {x\ y') and is such that the given point bisects the
part intercepted between the axes.
10. Find the equation to the straight line which passes through
the point ( - 4, 3) and is such that the portion of it between the axes
is divided by the point in the ratio 5 : 3.
Trace the straight lines whose equations are
11. a; + 2?/+3 = 0. 12. 5a--7//-9 = 0.
13. 3a; + 7r/ = 0. 14. 2a;-3?/ + 4 = 0.
Find the equations to the straight lines passing through the
following pairs of points.
15. (0, 0) and (2, -2). 16. (3, 4) and (5, 6).
17. (-1, 3) and (6, -7). 18. (0, -a) and (&, 0).
COORDINATE GEOMETRY.
[Exs. v.]
19. (a, &) and {a + h, a-h).
20. {at^, 2at-^) and (at^^ 2at;).
21. (a«„^)and(a«„^j.
22. (« cos 01 , a sin <pi) and (a cos (p^, a sin ^a)-
23. (acos0jLJ & sin 0j) and (acos02> ^sin^g)*
24. (* sec 01, 6 tan 0i) and (a sec 02, 6 tan 02).
Find the equations to the sides of the triangles the coordinates of
whose angular points are respectively
25. (1,4), (2,-3), and (-1,-2).
26. (0,1), (2,0), and (-1, -2).
27. Find the equations to the diagonals of the rectangle the
equations of whose sides are x = a, x = a\ y = b, and y = b\
28. Find the equation to the straight line which bisects the
distance between the points {a, b) and {a', b') and also bisects the
distance between the points ( - a, b) and (a', - b').
29. Find the equations to the straight lines which go through the
origin and trisect the portion of the straight line 3a; + 2/ = 12 which
is intercepted between the axes of coordinates.
Angles between straight lines.
66. To find the angle between two given straight lines.
Let the two straight lines be AL^ and AL^j meeting the
axis of X in L^ and L^,
I. Let their equations be
y - m^x^-G-^ and y ~ in.j,x ^r c.^
By Art. 47 we therefore have
tan^ZjA'^mi, and td^Vi. AL.^X^Wj.^,.
Now L L-^AL^^ - L AL^X - L AL.2.X.
tan L^AL^ - tan \AL^X - AL^X\
ta,n AL^X - tan AL^X rn^ - n^
1 + tan AL^X. tan AL^X
1 +mi«i2
ANGLES BETWEEN STRAIGHT LINES. 43
Hence the required angle - lL^AL
l + mim2
[In any numerical example, if the quantity (2) be a positive quan-
tity it is the tangent of the acute angle between the lines ; if negative,
it is the tangent of the obtuse angle.]
II. Let the equations of the straight lines be
^i£c + ^i2/ + Ci = 0,
and A^^x^- B^^y + G^^O.
By dividing the equations by B^ and B^, they may be
written
and
A,
c,
A.
L
sr
A'
Comparing
these with the equations of
(I-x
we
see
that
A
^2
Hence the required angle
- tan-i-, - i == tan ^
B
r(-J)
1 + mi7n.2 , / -41"^^ ^ -^2^
III. If the equations be given in the form
X cos a + y sin a - ^^ = 0 and x cos ^ + 2/ sin j^ - p-^ - O,
the perpendiculars from the origin make angles a and p
with the axis of x.
Now that angle between two straight lines, in which
the origin lies, is the supplement of the angle between the
perpendiculars, and the angle between these perpendiculars
is ^ - a.
|
[For, if OEi and OR2 be the perpendiculars from the origin upon
the two hnes, then the points O, R^, R^, and A lie on a circle, and
hence the angles R^OR^ and R.^AR^ are either equal or supplementary.]
44 COORDINATE GEOMETRY.
67. To find the condition that two straight lines may
he parallel.
Two straight lines are parallel when the angle between
them is zero and therefore the tangent of this angle is zero.
The equation (2) of the last article then gives
Two straight lines whose equations are given in the
"m" form are therefore parallel when their "7?i's" are the
same, or, in other words, if their equations differ only in
the constant term.
The straight line Ax + By + G' = 0 is any straight line which is
parallel to the straight line Ax + By + C = 0. For the "m's" of the
two equations are the same.
Again the equation A {x-x')+B {y-y') = 0 clearly represents the
straight line which passes through the point {x', y') and is parallel to
Ax + By + C=0.
The result (3) of the last article gives, as the condition
for parallel lines,
68. Ex. Find the equation to the straight line, which passes
through the point (4, - 5), and which is parallel to the straight line
3:c + 4r/ + 5--=0 (1).
Any straight line which is parallel to (1) has its equation of the
form
[For the "w" of both (1) and (2) is the same.]
This straight line will pass through the point (4, - 5) if
3x4 + 4x(-5) + C = 0,
i.e. if (7=20-12 = 8.
The equation (2) then becomes
3a;+42/ + 8 = 0.
69. To find the condition that two st^'aight lines j whose
equations are given, may he 'perpendicular.
Let the straight lines be
y - m^x -i-Ci,
and y - m.^x -\- G.2_.
CONDITIONS OF PERPENDICULARITY. 45
If 0 be the angle between them we have, by Art. 66,
1 +mim2
If the lines be perpendicular, then ^ = 90°, and therefore
tan 0 = 00 .
The right-hand member of equation (1) must therefore
be infinite, and this can only happen when its denominator
is zero.
The condition of perpendicularity is therefore that
1 + m^TTi^ - O, i.e. Tn-^Tn2 = - I.'
The straight line y - tu^x + c.^ is therefore perpendicular
y/c'-t
It follows that the straight lines
A^x +B^y + C^ = 0 and A^x + B^y + 0^ = 0,
for which m^ = - ^ and m^^ - ^ , are at right angles if
a) V A
i.e. a A^A^+B^B^ = 0.
70. From the preceding article it follows that the two
straight lines
A^x + B,y + Ci = Q (1),
and B,x-A,y+C^ = 0 (2),
are at right angles ; for the product of their m's
Also (2) is derived from (1) by interchanging the coefficients
of a; and y, changing the sign of one of them, and changing
the constant into any other constant.
Ex. The straight line through (x', y') perpendicular to (1) is (2)
where B^x' - A-^y' + 62= 0, so that Cg = A^y'- B^x'.
This straight line is therefore
B,{x-x')-A^{y-y') = 0.
46 COORDINATE GEOMETRY.
71. Ex. 1. Find the equation to the straight line which passes
through the point (4, - 5) and is perpendicular to the straight line
Sx + 4ij + 5 = 0 (1).
First Method. Any straight line perpendicular to (1) is by the
last article
4:X-Sij + C=0 (2).
[We should expect an arbitrary constant in (2) because there are
an infinite number of straight lines perpendicular to (1).]
The straight line (2) passes through the point (4, - 5) if
4x4-3x(-5) + C = 0,
i.e. a (7= -16-15= -31.
The required equation is therefore
4:X-Sy = 31.
Second Method. Any straight line passing through the given
point is
This straight line is perpendicular to (1) if the product of their
m's is - 1,
i.e. if m X ( - 1) = - 1,
i.e. if m=|.
The required equation is therefore
y + 5=i{x-4),
i.e. 4:X-'6y = Sl.
Third Method. Any straight line is y=mx + c. It passes through
the point (4, - 5), if
It is perpendicular to (1) if
mx{-i)=-l (4).
Hence m = f and then (3) gives c = - V.
The required equation is therefore y = '^x-^-^,
i.e. 4x-By = Sl.
|
[In the first method, we start with any straight line which is
perpendicular to the given straight line and pick out that particular
straight line which goes through the given point.
In the second method, we start with any straight line passing
through the given point and pick out that particular one which is
perpendicular to the given straight hne.
In the third method, we start with any straight line whatever and
determine its constants, so that it may satisfy the two given
conditions.
The student should illustrate by figures. ]
Ex. 2. Find the equation to the straight line which passes through
the point (x', y') and is perpendicular to the given straight line
yy' = 2a {x + x').
THE STRAIGHT LINE. 47
The given straight line is
yy' - 2ax - 2ax' = 0.
Any straight line perpendicular to it is (Art. 70)
2ay + xy'+G=0 (1).
This will pass through the point (x', y') and therefore will be the
straight line required if the coordinates x' and y' satisfy it,
i.eAt 2ai/ + xY+C = 0,
i.e. if G=-2ay' -x'y'.
Substituting in (1) for G the required equation is therefore
2a{y-y') + y'{x-x') = 0.
72. To find the equations to the straight lines which
pass through a given point (x', t/') and make a given angle a
with the given straight line y - nix + c.
Let P be the given point and let the given straight line
be LMJSf, making an angle 0
with the axis of x such that
tan 0 = m.
In general (i.e. except when
a is a right angle or zero) there
are two straight lines PMR and
FNS making an angle a with
the given line.
Let these lines meet the axis of x in R and S and let
them make angles ^ and <^' with the positive direction of
the axis of x.
The equations to the two required straight lines are
therefore (by Art. 62)
y -y' = tan ^ x (x - x) (1),
and y ~y' ~ t^^ <fi' X (x - x') (2).
Now cf, = L LMR + L RLM = a + 6*,
and <^'^L LNS+ L SLN= (180° ^a) + e.
Hence
■ , ^. tan a + tan ^ tana + m
i - tan a tan ff i - m tan a
and tan </>' = tan ( 1 80° + ^ - a)
. „ . tan d - tan a m - tan a
1 + tan 0 tan a 1 +7n tan a
48 COORDINATE GEOMETRY.
On substituting these values in (1) and (2), we have as
the required equations
^ ^ 1-mtana^ ^
^ ^ 1 + m tan a
EXAMPLES. VI.
Find the angles between the pairs of straight lines
1. x-ijsj^ - ^ and ^/3a;+2/ = 7.
2. ic-4?/ = 3 and ^x-y = ll. 3. 2/ = 3a3 + 7 and 3?/-a; = 8.
4. 2/ = (2-V3)a: + 5 and 2/= (2 + ^/3) a;- 7.
5. {m'^-mn)y = (inn^-n^)x + n^ and (?n7H- m^) y = (??i?i - w'^) a; + m^
6. Find the tangent of the angle between the lines whose inter-
cepts on the axes are respectively a, - 6 and 6, - a.
7. Prove that the points (2, - 1), (0, 2), (2, 3), and (4, 0) are the
coordinates of the angular points of a parallelogram and find the
angle between its diagonals.
Find the equation to the straight line
8. passing through the point (2, 3) and perpendicular to the
straight line 4a;-3i/ = 10.
9. passing through the point ( - 6, 10) and perpendicular to the
straight line 7aj + 8?/ = 5.
10. passing through the point (2, -3) and perpendicular to the
straight line joining the points (5, 7) and ( - 6, 3).
11. passing through the point (-4, -3) and perpendicular to the
straight line joining (1, 3) and (2, 7).
12. Find the equation to the straight line drawn at right angles to
the straight line - v = 1 through the point where it meets the axis
a b
of X.
13. Find the equation to the straight line which bisects, and is
perpendicular to, the straight line joining the points (a, b) and
14. Prove that the equation to the straight line which passes
through the point {a cos^ 6, a sin^ 6) and is perpendicular to the
straight line xsecd + y cosec d = ais x cos d-y sin d = a cos 26.
15. Find the equations to the straight lines passing through {x', y')
and respectively perpendicular to the straight lines
xxf-\-yy'=a\
[Exs. VI.]
EXAMPLES.
XX yy
a- 62
and x'y + xy' = a-.
|
16. Find the equations to the straight lines which divide, internally
and externally, the line joining (-3,7) to (5, - 4) in the ratio of 4 : 7
and which are perpendicular to this line.
17. Through the point (3, 4) are drawn two straight lines each
inclined at 45° to the straight line x-y = 2. Find their equations
and find also the area included by the three lines.
18. Shew that the equations to the straight lines passing through
the point (3, - 2) and incHned at 60° to the line
iJ3x + y = l are y + 2 = 0 and y -J3x + 2 + 3^S = 0.
19. Find the equations to the straight lines which pass through
the origin and are inclined at 75° to the straight line
20. Find the equations to the straight lines which pass through
the point {h, k) and are inclined at an angle tan~'^m to the straight
line y = mx + c.
21. Find the angle between the two straight lines 3a; = 4?/ + 7 and
5y = 12x + 6 and also the equations to the two straight lines which
pass through the point (4, 5) and make equal angles with the two
given lines.
73. To sheiv that the point (x', y') is on one side or the
other of the straight line Ax + By +(7 = 0 according as the
quantity Ax + By' + C is positive or negative.
Let LM be the given straight line and P any point
ix\ y).
Through P draw P^, parallel to
the axis of 3/, to meet the given
straight line in Q^ and let the co-
ordinates of Q be (.'«', y").
Since Q lies on the given line, we
have
Ax + C
y
i>x:x
so that
B
It is clear from the figure that PQ is drawn parallel to
the positive or negative direction of the axis of y according
as P is on one side, or the other, of the straight line LM^
i.e. according as y" is > or < y\
i.e. according as y" - y is positive or negative.
L.
50 COORDINATE GEOMETKY.
Now, by (1),
The point {x ^ y') is therefore on one side or the other of
LM according as the quantity Ax' + By' + C is negative or
positive.
Cor. The point (cc', y') and the origin are on the same
side of the given line if Ax + By + G and AxO-\- B xO + C
have the same signs, i.e. if Ax' + By' + C has the same sign
as C.
If these two quantities have opposite signs, then the
origin and the point (x, y') are on opposite sides of the
given line.
!74. The condition that two points may lie on the
same or opposite sides of a given line may also be obtained
by considering the ratio in which the line joining the two
points is cut by the given line.
For let the equation to the given line be
Ax + By-¥C=0 (1),
and let the coordinates of the two given points be [x-^, y^
and (a?2, 2/2)-
The coordinates of the point which divides in the ratio
7?ii : vu the line joining these points are, by Art. 22,
Tn^ + TYi.j 7)1-^ + m^ ^ ''
If this point lie on the given line we have
mj + mg m^ + m^ '
so that _i = _^ w^ - Ti (3 •
If the point (2) be between the two given points {x-^, y^)
and (x^, 2/2), i.e. if these two points be on opposite sides of
the given line, the ratio in-^ : 711^ is positive.
In this case, by (3) the two quantities Ax^ + By^ + C
and Ax^ + By^ + C have opposite signs.
The two points (a?i, y^ and {x^^ y^) therefore lie on the op-
LENGTHS OF PEKPENDICULARS. 51
posite (or the same) sides of the straight line Ax + By + (7 = 0
according as the quantities Ax^ + By^ + G and Ax.^ + By.^ + G
have opposite (or the same) signs.
Lengths of perpendiculars.
75. To find the length of the perpendicular let fcdl from,
a given point upon a given straight line.
(i) Let the equation of the straight line be
£CCOSa+ 2/ sin a - p = 0 (1),
so that, if p be the perpendicular on it, we have
ON^p and lXON=^o..
Let the given point F be {x ^ y').
Through P draw PR parallel to the given line to meet
OiV produced in R and draw PQ the required perpendicular.
If OR be 2^'i the equation to PR is, by Art. 53,
Since this passes through the point {x, y\ we have
£c' cos a + 2/' sin a. - p' = 0,
so that p = x' cos a + 2/' sin a.
But the required perpendicular
^PQ = NR = OR-ON = p'-p
= X' cos a + y' sin a - p (2).
|
The length of the required perpendicular is therefore
obtained by substituting x and y for x and y in the given
equation.
(ii) Let the equation to the straight line be
Ax + By + G=^0 (3),
the equation being written so that (7 is a negative quantity.
4 - 2
52 COORDINATE GEOMETRY.
As in Art. 56 this equation is reduced to the form (1)
by dividing it by V-^^ + B^. It then becomes
^A^-^B" slA^^B" ^|A^\E'
Hence
A , B ^ G
The perpendicular from the point {x\ y') therefore
= x' cos a + 2/" sin a - p
_ Ax' + By^ + C
VAZ + B2
The length of the perpendicular from (x, y') on (3) is
therefore obtained by substituting x and y for x and 2/ in
the left-hand member of (3), and dividing the result so
obtained by the square root of the sum of the squares of
the coefficients of x and y.
Cor. 1. The perpendicular from the origin
^G-rs]~AFVB\
Cor. 2. The length of the perpendicular is, by Art. 73,
positive or negative according as {x ^ y) is on one side or
the other of the given line.
76. The length of the perpendicular may also be
obtained as follows :
As in the figure of the last article let the straight line
meet the axes in L and M^ so that
OL^ -% and 0M=-%,.
A B
Let PQ be the perpendicular from F (re', y'^ on the
given line and PS and PT the perpendiculars on the axes
of coordinates.
"We then have
/\PML + /\MOL = /\OLP + AOPM,
i.e., since the area of a triangle is one half the product of
its base and perpendicular height,
PQ,LM+OL. OM^ OL.PS+ OM . PT.
EXAMPLES. 53
since (7 is a negative quantity.
Hence
Ax' + By' + 0
so that P<?r=
Ja^ + b^
EXAMPLES. VII.
Find the length of the perpendicular drawn from
1. the point (4, 5) upon the straight line 3a; + 4?/ = 10.
2. the origin upon the straight line -^ - j=l.
3. the point ( - 3, - 4) upon the straight line
■12{x + 6) = 5{y-2).
4. the point (&, a) upon the straight line - f =!•
5. Find the length of the perpendicular from the origin upon the
straight line joining the two points whose coordinates are
{a cos a, a sin a) and (a cos j8, a sin j8).
6. Shew that the product of the perpendiculars drawn from the
two points ( ± \/a2 - h^, 0) upon the straight line
- cos ^ + ^ sin ^= 1 is 62.
a 0
7. If p and p' he the perpendiculars from the origin upon the
straight lines whose equations are x sec ^ + ^ cosec 6 = a and
a; cos ^ - 2/ sin ^ = a cos 2^,
prove that 4Lp^+p'^ = a'^.
8. Find the distance between the two parallel straight lines
y = mx + c and y = mx + d.
9. What are the points on the axis of x whose perpendicular
X 1/
distance from the straight line - + ~ =lis a^
ah
10. Shew that the perpendiculars let fall from any point of the
straight line 2a; + 11?/ = 5 upon the two straight lines 24a; + 7t/ = 20
and 4a; -3?/ = 2 are equal to each other.
54 COORDINATE GEOMETRY. [EXS. VII.l
11. Find the perpendicular distance from the origin of the
perpendicular from the point (1, 2) upon the straight line
77. To find the coordinates of the foint of intersection
of two given straight lines.
Let the equations of the two straight lines be
a-^x + h{y + Ci = 0 (1),
and fta^^ + 522/ + ^2 = 0 {2)5
and let the straight lines be AL^ and AL^ as in the figure
of Art. 66.
Since (1) is the equation of AL^^ the coordinates of any
point on it must satisfy the equation (1). So the coordi-
nates of any point on AL^ satisfy equation (2).
Now the only point which is common to these two
straight lines is their point of intersection A.
The coordinates of this point must therefore satisfy
both (1) and (2).
If therefore A be the point {x^, t/i)? ^^ have
«ia^i + ^i2/i + Ci = 0 (3),
and a2^i + 522/i + <^2 = 0 W-
Solving (3) and (4) we have (as in Art. 3)
so that the coordinates of the required common point are
h-fi^ - h.^c^ - c^a^ - c^a^
78. The coordinates of the point of intersection found
in the last article are infinite if
a^^ - a^hi = 0.
But from Art. 67 we know that the two straight lines
are parallel if this condition holds.
|
Hence parallel lines must be looked upon as lines whose
point of intersection is at an infinite distance.
CONCUKEENCE OF STRAIGHT LINES. 55
79. To find the condition that three straight lines may
ineet in a point.
Let their equations be
a^x + hyy + c^ = 0 (1),
a.jX + h^y + C2 = 0 (2),
and a.p: + h-^y + C3 = 0 (3).
By Art. 77 the coordinates of the point of intersection
of (1) and (2) are
- r -J- and - jf J-- (4).
If the three straight lines meet in a point, the point of
intersection of (1) and (2) must lie on (3). Hence the
values (4) must satisfy (3), so that
0-iCn - Oc\C-\ -t C-\Cto C.-^Cti -
i. e. a^ (b^c^ - b^c-^) + 63 (ci^g - 02%) + c^ (a^b^ - a^bj) - 0,
i. e. ai (b^c^ - b^c^) + b^ {c^a^ - c^a^ + Cj (^2^3 - ^3^2) = 0 . . , (5).
Aliter. If, the three straight lines meet in a point let
it be (ccj, 2/1), so that the values x\ and y-^ satisfy the
equations (1), (2), and (3), and hence
a-^x-^ + 6i2/i + Ci = 0,
ftoa^i + b^yi + C2 = 0,
and a^x-^ + b^y^ + Cg = 0.
The condition that these three equations should hold
between the two quantities x^ and y^ is, as in Art. 12,
which is the same as equation (5).
80. Another criterion as to whether the three straight
lines of the previous article meet in a point is the following.
If any three quantities ^p, q, and r can be found so
that
p (ajX + b^y + C;^) + q (a^x + b.2y + c^) + r (a^x + b^y + Cg) = 0
identically, then the three straight lines meet in a point.
56 COORDINATE GEOMETRY.
For in this case we have
a^x + h^y + Cg =: - - (a-^x + h^y + Cj) - - {a^x + h^y + Co) . • •(!).
Now the coordinates of the point of intersection of the
first two of the lines make the right-hand side of (1) vanish.
Hence the same coordinates make the left-hand side vanish.
The point of intersection of the first two therefore satisfies
the equation to the third line and all three therefore meet
in a point.
81. Ex. 1. Shew that the three straight lines 2«-3i/ + 5 = 0,
dx + ^y -1 = 0, and 9a; -5y 4- 8 = 0 meet in a point.
If we multiply these three equations by 6, 2, and - 2 we have
identically
6 (2x - 32/ + 5) + 2 (3x -I- 4t/ - 7) - 2 (9a; - 5?/ -F 8) = 0.
The coordinates of the point of intersection of the first two lines
make the first two brackets of this equation vanish and hence make
the third vanish. The common point of intersection of the first two
therefore satisfies the third equation. The three straight lines
therefore meet in a point.
Ex. 2. Prove that the three 'perpendiculars draicn from the
vertices of a triangle upon the opposite sides all meet in a point.
Let the triangle be ABC and let its angular points be the points
K,2/i). (^2 '2/2)5 and (x^^ys)-
The equation to BG is y-y^ = "^ - - {x - x^).
The equation to the perpendicular from A on this straight line is
x^ x^
i'^' 2/ (2/3 -2/2) + ^(^3 -^2) =2/1 (1/3 -2/2) +^1(^3 -^2) (!)•
So the perpendiculars from B and C on CA and AB are
2/(2/i-2/3)+^(aa-^3) = 2/2(2/i-2/3)+«'2{^i-^3) (2).
and 2/ (2/2 -2/]) + ^(^2-^1) =2/3 (2/2 -2/1) + ^3 (^2-^1) (3).
On adding these three equations their sum identically vanishes.
The straight lines represented by them therefore meet in a point.
This point is called the ortliocentre of the triangle.
82. To Jind the equation to any straight line which
2)asses through the intersection of the two straight lines
and a^x + b^y + Cg = 0 (2).
INTERSECTIONS OF STRAIGHT LINES. 57
If (a?i, 2/1) be the common point of the equations (1)
and (2) we may, as in Art. 77, find the values of x^ and 2/1,
and then the equation to any straight line through it is
where m is any quantity whatever.
Aliter. If A be the common point of the two straight
lines, then both equations (1) and (2) are satisfied by the
coordinates of the point A.
Hence the equation
a-^x + h^y + Ci + X {a,^x + h^y + Co) = 0 (3)
is satisfied by the coordinates of the common point A,
where \ is any arbitrary constant.
But (3), being of the first degree in x and y, always
represents a straight line.
|
It therefore represents a straight line passing through A .
Also the arbitrary constant X may be so chosen that (3)
may fulfil any other condition. It therefore represents
any straight line passing through A.
83. Ex. Find the equation to the straight line ivhich passes
through the intersection of the straight lines
2x-Sy + 4: = 0, Sx + ^y-5 = 0 (1),
and is perpendicular to the straight line
Gx-7y + 8 = 0 (2).
Solving the equations (1), the coordinates x^^, 7/1 of their common
point are given by
^1 ^ yi _ 1 _ 1
(-3)(-5)-4x4 4x3-2x(-5) 2x4-3x(-3) ^^'
so that x-^= - yY and 2/1 =Tf-
The equation of any straight line through this common point is
therefore
This straight line is, by Art. 69, perpendicular to (2) if X f = - 1, i.e. if mr= - |.
The required equation is therefore
i.e. 119a; + 1022/ = 125.
COORDINATE GEOMETRY.
Aliter. Any straight line through the intersection of the straight
lines (1) is
i.e. (2 + 3X)a; + ?/(4X-3) + 4-5X = 0 (3).
This straight line is perpendicular to (2), if
6(2 + 3X)-7(4X-3) = 0, (Art. 69)
The equation (3) is therefore
i.e. 119a; +102?/ -125 = 0.
Bisectors of angles between straight lines.
84. To find the equations of the bisectors of the angles
between the straight lines
ajpc + b-yy + Cj^ = 0 (1),
and a^ + b^ + c^ = 0 (2).
Let the two straight lines be AL^ and AL^, and let the
bisectors of the angles between them be AM^ and A^f^-
Let P be any point on either of these bisectors and
draw PiVi and FJV^ perpendicular to the given lines.
The triangles PAN-^ and PAN<^ are equal in all respects,
so that the perpendiculars PN^ and PN^ ^-re equal in
magnitude.
Let the equations to the straight lines be written
so that Ci and c^ are both negative, and to the quantities
Ja^ + b^ and Ja^ + b^ let the positive sign be prefixed.
EQUATIONS TO BISECTOKS OF ANGLES. 59
If P be the point (/i, k), the numerical values of PN^
and PN^ are (by Art. 75)
aJh + hJc + c. , aji + hjc + c^ , ,
If P lie on AM^^ i.e. on the bisector of the angle
between the two straight lines in which the origin lies, the
point P and the origin lie on the same side of each of the
two lines. Hence (by Art. 73, Cor.) the two quantities (1)
have the same sign as c^ and c^ respectively.
In this case, since c^ and c^ have the same sign, the
quantities (1) have the same sign, and hence
aji + hjc + Ci aji + hjc + c^
But this is the condition that the point (h, k) may lie on
the straight line
a^x + h-{y + Cj a^ + h^y + c^
which is therefore the equation to AM-^.
If, however, P lie on the other bisector AM^, the two
quantities (1) will have opposite signs, so that the equation
to AM^ will be
a-^x + \y + Ci a^x + })c^y + c^
The equations to the original lines being therefore
arranged so that the constant terms are both positive (or
both negative) the equation to the bisectors is
VS7+b7 - Vi7+bp '
the upper sign giving the bisector of the angle in which
the origin lies.
85. Ex. Find the equations to the bisectors of the angles
between the straight lines
3x-4:y + 7 = 0 and 12x-5y-S = 0.
Writing the equations so that their constant terms are both
positive they are
Sx-Ay + 7 = 0 and -12x + 5y + 8 = 0.
COORDINATE GEOMETRY.
t.e.
i.e.
The equation to the bisector of the angle in which the origin lies
is therefore
Bx-^y + 7 _-12x + 5y + 8
i.e. 13{Bx-^y + l) = 5{-12x + 5y + 8),
i.e. 99a; -772/ + 51 = 0.
The equation to the other bisector is
Sx-iy + l _ - 12x + 5?/ + 8
V32 + 42 ~ ~ x/122 + 52 '
21a: + 272/- 131 = 0.
86. It will be found useful in a later chapter to have
the equation to a straight line, which passes through a
given point and makes a given angle 6 with a given line, in
a form different from that of Art. 62.
Let A be the given point {h, k) and L'AL a straight
line through it inclined at an
angle 6 to the axis of x.
Take any point F, whose
coordinates are (x, y), lying on
this line, and let the distance
AP be r.
Draw PM perpendicular
to the axis of x and AN perpendicular to PM.
|
Then x--h=- AN = AP cosO=r cos 6,
and y - k = NP - AP sin ^ = r sin 0.
x-h y-k
Hence
This being the relation holding between the coordinates
of any point P on the line is the equation required.
Cor. From (1) we have
x - h + r cos 6 and y = k + r sin 0.
The coordinates of any point on the given line are
therefore h-\-r cos 6 and k + r sin 0.
87. To find the length of the straight line drawn
through a given point in a given direction to meet a given
straight line.
EXAMPLES. 61
Let the given straight line be
Ax-¥By + C^Q (1).
Let the given point A be (7t, k) and the given direction
one making an angle 6 with the axis of x.
Let the line drawn through A meet the straight line
(1) in P and let AP be r.
By the corollary to the last article the coordinates
of P are
h + r cos 6 and k + r sin 6.
Since these coordinates satisfy (1) we have
Ah-[-Bk+C
A cos 6 ■¥ B sin 6
giving the length AP which is required.
Cor. From the preceding may be deduced the length
of the perpendicular drawn from (A, k) upon (1).
For the "m" of the straight line drawn through A is
tan ^ and the "m" of (1) is - -.
This straight line is perpendicular to (1) if
i. e. if tan ^ = - ■ ,
so that
and hence
A
cos 6 sin 6 1
■4 B JA'- + B'
AcosO + BsinO= 4=^£L^s/A^ + B'.
Substituting this value in (2) we have the magnitude
of the required perpendicular.
EXAMPLES. VIII.
Find the coordinates of the points of intersection of the straight
lines whose equations are
1. 2x-Si/ + 5 = 0 and 7x-¥^y=S.
62 COOEDINATE GEOMETRY. [EXS.
a 0 ha
3 a ^ a
. y = vi-,x-\ and y=moX-\ .
4. a; cos 01 + ^ sin 01 = a and a; cos 02+2/ sin 02 = «•
5. Two straight lines cut the axis of x at distances a and - a and
the axis of y at distances & and 6' respectively ; find the coordinates
of their point of intersection.
6. Find the distance of the point of intersection of the two
straight lines
2x-Sy + 5 = 0 and dx + 4:y = 0
from the straight line
5x-2y = 0.
7. Shew that the perpendicular from the origin upon the
straight line joining the points
(a cos a, a sin a) and {a cos /3, a sin ^)
bisects the distance between them.
8. Find the equations of the two straight lines drawn through
the point (0, a) on which the perpendiculars let fall from the point
(2a, 2a) are each of length a.
Prove also that the equation of the straight line joining the feet
of these perpendiculars is y -r2x = oa.
9. Find the point of intersection and the inclination of the two
lines
Ax + By = A+B and A{x-y)+B{x + y)=2B.
10. Find the coordinates of the point in which the line
2y-Sx + l = 0
meets the line joining the two points (6, - 2) and ( - 8, 7). Find also
the angle between them.
11. Find the coordinates of the feet of the perpendiculars let fall
from the point (5, 0) upon the sides of the triangle formed by joining
the three points (4, 3), (-4, 3), and (0, -5); prove also that the
points so determined lie on a straight line.
12. Find the coordinates of the point of intersection of the
straight lines
2x-3y=^l and 5y-x = S,
and determine also the angle at which they cut one another.
13. Find the angle between the two lines
Sx + y + 12 = 0 and x + 2y-l = 0.
Find also the coordinates of their point of intersection and the
equations of lines drawn perpendicular to them from the point
VIII.] EXAMPLES. 63
14. Prove that the points whose coordinates are respectively
(5, 1), (1, -1), and (11, 4) lie on a straight line, and find its intercepts
on the axes.
Prove that the following sets of three lines meet in a point.
15. 2x-Sy = 7, Sx-4:y = 13, and 8x-lly = S3.
16. dx + 4.y + G = 0, 6x + 5y + 9 = 0, and Sx + Sy + 5 = 0.
17. - + 7 = 1, j+^ = l, and y = x.
abba
18. Prove that the three straight lines whose equations are
15a;- 18?/ + 1 = 0, 12x + lOi/ - 3 = 0, and 6x + QQy-ll = 0
all meet in a point.
Shew also that the third line bisects the angle between the other
two.
19. Find the conditions that the straight lines
y = m-^x + ai, y = m^-\-a^, and y = m2X-\-a^
|
may meet in a point.
Find the coordinates of the orthocentre of the triangles whose
angular points are
20. (0,0), (2, -1), and (-1,3).
21. (1,0), (2,-4), and (-5,-2).
22. In any triangle ABG^ prove that
(1) the bisectors of the angles A, B, and C meet in a point,
(2) the medians, i.e. the lines joining each vertex to the middle
point of the opposite side, meet in a point,
and (3) the straight lines through the middle points of the sides
perpendicular to the sides meet in a point.
Find the equation to the straight line passing through
23. tlie point (3, 2) and the point of intersection of the lines
2x + Sy = l and Sx-Ay = Q.
24. the point (2, - 9) and the intersection of the lines
2x + 5y-8 = 0 and 3x-4y=^S5.
25. the origin and the point of intersection of
x~y-4i=0 and lx + y + 20=0,
proving that it bisects the angle between them.
26. the origin and the point of intersection of the lines
- + f = 1 and Y + ^ = 1. b b a
27. the point (a, b) and the intersection of the same two lines.
28. the intersection of the lines
x-2y-a=0 and x + 3y-2a = 0
64 COORDINATE GEOMETRY. [Exs.
and parallel to the straight line
29. the intersection of the lines
x + 2y + S = 0 and 3x + iy + 7 = 0
and perpendicular to the straight line
y-x = 8.
30. the intersection of the lines
dx-iy + l = 0 and 5x + y -1=0
and cutting off equal intercepts from the axes.
31. the intersection of the lines
2x~By = 10 and x + 2y:=Q
and the intersection of the lines
16a;-102/ = 33 and 12x + Uy + 29 = 0.
32. If through the angular points of a triangle straight lines be
drawn parallel to the sides, and if the intersections of these Hnes be
joined to the opposite angular points of the triangle, shew that the
joining lines so obtained will meet in a point.
33. Find the equations to the straight lines passing through the
point of intersection of the straight lines
Ax + By + C = 0 and A'x + B'y + C'^0 and
(1) passing through the origin,
(2) parallel to the axis of y,
(3) cutting off a given distance a from the axis of y,
and (4) passing through a given point {x', y').
34. Prove that the diagonals of the parallelogram formed by the
four straight lines
^?>x + y = 0, ^?>y + x=^0, jBx + y = l, and JBy + x = l
are at right angles to one another.
35. Prove the same property for the parallelogram whose sides
are
a b 0 a a 0 o a
36. One side of a square is inclined to the axis of x at an angle a
and one of its extremities is at the origin ; prove that the equations
to its diagonals are
y (cos a - sin a) = x (sin a + cos a)
and ?/ (sin a + cos a) + a: (cos a -sin a) = a.
Find the equations to the straight lines bisecting the angles
between the following pairs of straight lines, placing first the bisector
of the angle in which the origin lies.
37. x+ysJB = %-\-2JB and a;-?/ ^3 = 6-2^3.
VIII.] EXAMPLES. 65
38. 12x + 5y-4c = 0 and Bx+4.tj + 7 = 0.
39. ix + Sij -7 = 0 and 24^ + 7ij- 31 = 0.
40. 2x + y=4: and y + Sx = 5.
41. y-b=^ i>(^-«) and y-h = z ^(rc-a).
Find the equations to the bisectors of the internal angles of the
triangles the equations of whose sides are respectively
42. 3x + 4:y=e, 12x-5y=B, and 4:X-3y + 12 = 0.
43. Sx + 5y=15, x + y=4:, and 2x + y = Q.
44. Find the equations to the straight lines passing through the
foot of the perpendLcular from the point {h, Jc) upon the straight line
Ax + By + G = 0 and bisecting the angles between the perpendicular
and the given straight line.
45. Find the direction in which a straight Kne must be drawn
through the point (1, 2), so that its point of intersection with the line
x + y = 4: may be at a distance ^^6 from this point.
CHAPTER V.
THE STRAIGHT LINE {continued).
POLAR EQUATIONS. OBLIQUE COORDINATES.
MISCELLANEOUS PROBLEMS. LOCI.
88. To find the general equation to a straight line in
polar coordinates.
Let p be the length of the perpendicular 0 Y from the
origin upon the straight line, and
let this perpendicular make an
angle a with the initial line.
|
Let P be any point on the
line and let its coordinates be r
and 6.
The equation required will
then be the relation between r, 6, p, and a.
From the triangle 0 YP we have
p = r cos YOP = rcos{a-6)=^r cos (6 - a).
The required equation is therefore
r cos (6 - a) =p.
[On transforming to Cartesian coordinates this equation becomes
the equation of Art. 53.]
89. To find the polar equation of the straight line
joining the poiiits whose coordinates are (r^, 6^) and {r^, 6^.
THE STRAIGHT LINE. OBLIQUE COORDINATES. 67
Let A and B be the two given points and P any point
on the line joining them
whose coordinates are r and
e.
Then, since
AA0B-=AA0P+A POB,
we have
I.e.
J r^r^ sin A OB = J r^r sin AOP + ^ ri\ sin POB,
r^r^ sin {6 2 - 6^ = r^r sin {0 - 6-^ + rr^ sin (0^ - 6),
sin (^,-^i)_ sin ((9-^1) sin ((92-^)
OBLIQUE COORDINATES.
90. In the previous chapter we took the axes to be
rectangular. In the great majority of cases rectangular
axes are employed, but in some cases oblique axes may be
used with advantage.
In the following articles we shall consider the proposi-
tions in which the results for oblique axes are different
from those for rectangular axes. The propositions of Arts.
50 and 62 are true for oblique, as well as rectangular,
coordinates.
91. To find the equation to a straight line referred to
axes inclined at an angle w.
Let LPL' be a straight line which cuts the axis of
a distance c from the origin and is
inclined at an angle 0 to the axis
of X.
Let P be any point on the
straight line. Draw PNM parallel
to the axis of y to meet OX in M^
and let it meet the . straight line
through C parallel to the axis of x
in the point N,
Let P be the point (.r, y\ so that
CN^OM^x, and NP = MP- 00 ^y-c.
5-
Fat
68 COORDINATE GEOMETRY.
Since L GPN= l FNN' - l PCN' = w - ^, we have
y-c NP _B>inNCP _ sinO
~ir ~ 'CN~ ^^PN~ sin (o)- ^) •
TT Si^^ /1\
Hence y = x-. - ; 7:. + c (1).
3?his equation is of the form
y = mx + c,
where
sin^ sin^ tan^
7)1 =
sin (o) - 6) sin w cos 0 - cos to sin 0 sin <o - cos w tan ^ '
, , „ J /> ^ sin CD
and thererore tan v = .
1 + 7)1 cos a>
In oblique coordinates the equation
y = mx + c
therefore represents a straight line which is inclined at an
angle
- m sin o)
tan~i
l+mcosco
to the axis of x.
Cor. From. (1), by putting in succession 0 equal to 90°
and 90° + o), we see that the equations to the straight lines,
passing through the origin and perpendicular to the axes of
X and y, are respectively y = and y = - x cos w.
92. The axes being oblique, to find the equation to the
straight line, such that the 'perpendicular on it from the origin
is of length p and makes angles a and ^ with the axes of x
and y.
Let LM be the given straight line and OK the perpen-
dicular on it from the origin.
Let P be any point on the
straight line ; draw the ordinate
PN and draw NP perpendicular
to OK and PS "oerpendicular to
NR.
Let P be the point {x, y), so
that OJV = X and NP = y.
THE STRAIGHT LINE. OBLIQUE COORDINATES. 69
The lines NP and 0 Y are parallel.
Also OK and SP are parallel, each being perpendicular
to NB.
Thus lSPN^lKOM=^.
We therefore have
'p = OK- OR + SP = OiVcos a + NP cos p=^xcosa + y cos ^.
Hence x cos a + y cos /5 - p = 0,
being the relation which holds between the coordinates of
any point on the straight line, is the required equation.
93. To find the angle between the straight lines
y = mx + c and y = mx + c,
the axes being oblique.
If these straight lines be respectively inclined at angles
0 and 0' to the axis of x, we have, by the last article,
^ msinoD , ^ ,,, m'sinw
tan u = :. and tan u -
1 + ni cos CO 1 + 711 cos 0)
The angle required is 0~ 0'.
XT J. ir\ t\t\ tan ^- tan ^'
Now tan(e-e)=j^:^^^^^-^^,
m sm CD in sm co
1 +m cos CO 1 + 771 cos CO
m sin CO m' sin co
1+771 cos CO 1 + m' cos CO
_ m sin CO ( 1 + m' cos co) - m' sin co (1 + 7?i. cos co)
(1 + m cos co) (1 + W cos co) + ?92m' sin^ co
_ (m - w') sin CO
|
1 + (m + m ) cos CO + mm' '
The required angle is therefore
tan
_j (m - in!) sin co
1 + {m + TTb) cos 0) + 7f}im' '
Cor. 1. The two given lines are parallel if m = m'.
Cor. 2. The two given lines are perpendicular if
1 + (m + m') cos 0} + mm' = O.
70 COORDINATE GEOMETRY.
94. If the straight lines have their equations in the
form
Ax-\-By + G = 0 and A'x + B'y ■\- C = 0,
then 7n = - ^ and m =--Wt'
Substituting these values in the result of the last article
the angle between the two lines is easily found to be
J A'B-AB' .
AA' + BB' - {AB' + A'B) cos w
The given lines are therefore parallel if
A'B-AB'=^0.
They are perpendicular if
AA' + BB' = {AB' + A'B) cos w.
95. Ex. The axes being inclined at an angle of 30°, obtain the
equations to the straight lines ivhich pass through the origin and are
inclined at 45° to the straight line x + y = l.
Let either of the required straight lines be y = mx.
The given straight line isy= -x + 1, so that m'= - 1.
We therefore have
1 + (m + m') cos <a + mm'
where m'= - 1 and w = 30°.
This equation gives 2 + (,,,_ 1)^3- 2m= "^^^
Taking the upper sign we obtain m= - j^.
Taking the lower sign we have m = - ^3.
The required equations are therefore
y=-sjSx and y=z--^x,
i.e. y + iJ3x=^0 and JSy + x = 0.
96. To find the length of the lyerpendicular froin the
point (x'j y) upon the straight line Ax + By +0 - 0, the axes
being inclined at an angle <a, and the equation being written
so that C is a negative quantity.
THE STRAIGHT LINE. OBLIQUE COORDINATES. 71
Let the given straight line meet the axes in L and M^
G C
so that OL = --, and OM^ - -=, .
A B
Let P be the given point {x', y').
Draw the perpendiculars PQ, PR,
and PS on the given line and the
two axes.
Taking 0 and P on opposite sides
of the given line, we then have
IiLPM + AMOL=aOLP + aOPM,
i.e. PQ . LM + OL . OM silicon OL . PE + OM . PS. ..{I).
Draw PU and PV parallel to the axes of y and x, so
that PU = y' and PV-^x.
Hence PE ^ PU sin PUR = y' sin w,
and PS =PV sin P VS - x sin w.
Also
LM= s/OL^ + OM^ - 20 L . Oif cos w
V
(72 C'
0^
- (7 /l+l
2 cos w
since C is a negative quantity.
On substituting these values in (1), we have
pc><(-o)xy-i+-i
2 cos CO C^ .
c
c
so that
PQ =
--7.7/ sm o> - Pi . £c sm 00,
Ax^ + By' + C
VA2 + B2 - 2 AB cos CO
. sm a;.
Cor. If (0 - 90°, i.e. if the axes be rectangular, we
have the result of Art. 75.
72 COORDINATE GEOMETRY.
EXAMPLES. IX.
1. The axes being inclined at an angle of 60°, find the inclination
to the axis of x of the straight lines whose equations are
and (2) 2y=.{^^-l)x + l.
2. The axes being inclined at an angle of 120°, find the tangent
of the angle between the two straight lines
Qx + ly=.l and 28a; - 73?/ = 101.
3. With oblique coordinates find the tangent of the angle
between the straight lines
y = mx + c and my+x = d.
4. liy=x tan - - - and y=x tan - j represent two straight lines
at right angles, prove that the angle between the axes is ~ .
5. Prove that the straight lines y + x=c and y=x + d are at
right angles, whatever be the angle between the axes.
6. Prove that the equation to the straight line which passes
through the point {h, Jc) and is perpendicular to the axis of x is
x + y cos o) = h+k cos w.
7. Find the equations to the sides and diagonals of a regular
hexagon, two of its sides, which meet in a corner, being the axes of
coordinates.
8. From each corner of a parallelogram a perpendicular is drawn
upon the diagonal which does not pass through that corner and these
are produced to form another parallelogram ; shew that its diagonals
are perpendicular to the sides of the first parallelogram and that they
both have the same centre.
9. If the straight lines y = miX + Cj^ and y=m,^x + C2 make equal
angles with the axis of x and be not pariallel to one another, prove
that iiij^ + ^2 + 2mjin2 cos w = 0.
|
10. The axes being inclined at an angle of 30°, find the equation
to the straight line which passes through the point ( - 2, 3) and is
perpendicular to the straight line y + Bx = 6.
11. Find the length of the perpendicular drawn from the point
(4, -3) upon the straight line 6a; + 3^ -10 = 0, the angle between the
axes being 60°.
12. Find the equation to, and the length of, the perpendicular
drawn from the point (1, 1) upon the straight line 3a; + 4?/ + 5 = 0, the
angle between the axes being 120°.
[EXS. IX.] THE STRAIGHT LINE. PROBLEMS. 73
13. The coordinates of a point P referred to axes meeting at an
angle w are [h, k) ; prove that the length of the straight line joining
the feet of the perpendiculars from P upon the axes is
sin w ^y/i^ +k- + 2hk cos w.
14. From a given point {h, k) perpendiculars are drawn to the
axes, whose inclination is oj, and their feet are joined. Prove that
the length of the perpendicular drawn from [h, Jc) upon this line is
hk sin^ 0}
JhF+W+2hkcos^'
and that its equation is hx - ky = h^- k^.
Straight lines passing through fixed points.
97. If the equation to a straic/ht line be of the form
where \ is any arbitrary constant^ it always passes through
one fixed point whatever be the value of \.
For the equation (1) is satisfied by the coordinates of
the point which satisfies both of the equations
ax + by + G~Oj
and a'x 4- b'y + c' = 0.
This point is, by Art. 77,
'be' - b'c ca' - c'a^
^ab' - a'b ' ab' - a'b/
and these coordinates are independent of A.
Ex. Given the vertical angle of a triangle in magnitude and
position, and also the sum of the reciprocals of the sides ivhich contain
it; shew that the base always passes through a fixed point.
Take the fixed angular point as origin and the directions of the
sides containing it as axes ; let the lengths of these sides in any such
triangle be a and &, which are not therefore given.
We have - + 7=const. = r- (say) (1).
do K
The equation to the base is
a 0
1 y
COORDINATE GEOMETRY.
"Whatever be the value of a this straight line always passes through
the point given by
x-y = 0 and |-1 = 0,
i.e. through ih.e fixed point {k, k).
k
98. Prove that the coordinates of the centre of the
circle inscribed in the triangle, whose vertices are the points
(^ij 2/i)> (^2 J 2/2)5 «^^G? (a^3, 2/3), are
axi + hx^ + cx^ ay^ + hy.2 + cy^
a+b+G a+b+c '
where a, 6, «7ic? c are ^Ae lengths of the sides of the triangle.
Find also the coordinates of the centres of the escribed
circles.
Let ABC be the triangle and let AD and CE be the
bisectors of the angles A and G
and let them meet in 0'.
Then 0' is the required point.
Since AD bisects the angle
BAC we have, by Euc. YI. 3,
^ _DG_ BD + DG _ _a_
BA ~ AG~ BA+'AC~bTc'
so that
ba
(xj.yj)
DG
b + c
Also, since GO' bisects the angle AGD, we have
0'D~ CD~ ba
a
The point D therefore divides BG in the ratio
BA : AG, i.e. c : b.
Also 0' divides AD in the ratio b + c : a.
Hence, by Art. 22, the coordinates of D are
cxs + bx^ fi ^3 + ^.'/2
c + b
c + b
THE STRAIGHT LINE. PROBLEMS.
Also, by the same article, the coordinates of 0' are
cxo + hxo ,, X cVo + hy.^
and
a
ax^ + bx^, + Gx.^ - ay^ 4- by^ + cy^
a + b + G a + b + c
Again, if 0^ be the centre of the escribed circle opposite
to the angle -4, the line COi bisects the exterior angle of
ACB.
Hence (Euc. VI. A) we have
AO, _AC__ b^G
Therefore Oi is the point which divides AD externally in
the ratio b + g : a.
Its coordinates (Art. 22) are therefore
,y . GXo + bx^ ,-, , cy., + by»
ana
a
- axi + bx^ + Gx^ - - ayi + by. 2 4- Gy^
- a+b+G - a+b+c
Similarly, it may be shewn that the coordinates of the
escribed circles opposite to B and C are respectively
^aXj^ - bx2 + GXs ay^ - by^ + cyA
c-
a - b+G ' a - b+c
and /axj^x2-_cxs ay^ + by.^ - Gy^
99. As a numerical example consider the case of the
triangle formed by the straight lines
3x+4cy-7 = 0, 12x + 6y-l7 =0 and 5x + 12y - 34: = 0.
|
These three straight lines being BC, CA, and AB
respectively we easily obtain, by solving, that the points
A, B, and C are
76 COORDINATE GEOMETRY,
Hence
682 5p
17 ,j.-^, 85
and
?V /^l_i?V- A' 12-'_ 13
/ 72 52Y /19 _ 62Y _ /395T
V V7 "^ 16/ "^ VT~ W ~ V "~Tl2
165^
^^Vl69 '^'
112" 112'
Hence
85 2 170 85 19 1615
13 -52__676 13 67 871
429 , 429
c^z = YY9,> and C2/3 = Yj2-
The coordinates of the centre of the incircle are therefore
170 _ 6^6 m 1615 871 429
1X2 ~ iT2 "^ 112 , IT2" "^ 112 '^ II2
85 13 429 85 13 429'
16 ^T"^ 112 16 "^y^ 112
-1 ,265
^ and ^.
The length of the radius of the incircle is the perpen-
dicular from ( - T^ , jYo ) ^P^^ *^® straight line
3aj + 42/ - 7 = 0,
THE STRAIGHT LINE. PROBLEMS. 77
and therefore ^
i^)^0
-21 + 1060-784 255 51
5x112 5x112 112'
The coordinates of the centre of the escribed circle
which touches the side BG externally are
_170_676 429 1615 871 429
112 112^112 ~Tl2" "^ 112"^Tl2
_85 13 429 85 13 429 '
"le'^T'^m ~l6'^T'^ri2
-417 , -315
Similarly the coordinates of the centres of the other
escribed circles can be written down.
100. Ex. Find the radius, and the coordinates of the centre, of
the circle circumscribing the triangle formed by the points
(0, 1), (2, 3), and (3, 5).
Let (iCi , 2/j) be the required centre and R the radius.
Since the distance of the centre from each of the three points is the
same, we have
^i'+ (2/1 - 1)'= (^1 - 2)H (^/i - 3)2= (.ri - 3)2+ (y, - 5)2= J22...(l).
From the first two we have, on reduction,
From the first and third equations we obtain
Solving, we have x^= - ^ and yi=^.
Substituting these values in (1) we get
i2=tx/10.
101. Ex. Prove that the middle points of the diagonals of a com-
plete quadrilateral lie on the same straight line.
[Complete quadrilateral. Def. Let OAGB be any quadrilateral.
Let AG and OB be produced to meet in E, and BG and OA to meet in
F. Join AB, OC, and EF. The resulting figure is called a complete
quadrilateral ; the lines AB,OG, and EF are called its diagonals, and
the points E, F, and D (the intersection of AB and OC) are called its
vertices.]
78 COOEDINATE GEOMETRY.
Take the lines OAF and OBE as the axes of x and y.
1Y
BI ^
n^X
[A
^^^ A F X
Let 0A = 2a and 0B = 2b, so that A is the point (2a, 0) and B is
the point (0, 2b); also let C be the point {2h, 2k).
Then L, the middle point of OC, is the point {h, h), and ilf, the
middle point of AB, is (a, fo).
The equation to LM is therefore
£.e. {Ji-a)y-{k-h)x = bh-ak (1).
k-h
Again, the equation to BC is y -2b= - ~x.
Putting ?/ = 0, we have x = ^ - - , so that F is the point
f 2ak \
Similarly, E is the point ( 0, - _ j .
Hence N, the middle point of EF^ is f r; - r- , ■ j .
These coordinates clearly satisfy (1), i.e. N lies on the straight
line L3I.
EXAMPLES. X.
1. A straight line is such that the algebraic sum of the perpen-
diculars let fall upon it from any number of fixed points is zero;
shew that it always passes through a fixed point.
2. Two fixed straight lines OX and 0 Y are cut by a variable line
in the points A and B respectively and P and Q are the feet of the
perpendiculars drawn from A and B upon the lines OBY and OAX.
Shew that, if ^B pass through a fixed point, then PQ will also pass
through a fixed point.
[EXS. X.] THE STRAIGHT LINE. PROBLEMS. 79
3. If the equal sides AB and AC of au isosceles triangle be pro-
duced to E and F so that BE .GF = AB% shew that the line EF will
always pass through a fixed point.
4. If a straight line move so that the sum of the perpendiculars
let fall on it from the two fixed points (3, 4) and (7, 2) is equal to
three times the perpendicular on it from a third fixed point (1, 3),
prove that there is another fixed point through which this line always
passes and find its coordinates.
Find the centre and radius of the circle which is inscribed in the
triangle formed by the straight lines whose equations are
5. 3a; + 42/ + 2 = 0, 3x-4.y + 12 = 0, and 4x-3y = 0.
|
6. 2x + iy + S = 0, 4x + Sy + 3 = 0, and a;+l = 0.
7. y = 0, 12x-5y=0, and 3a; + 4?/-7 = 0.
8. Prove that the coordinates of the centre of the circle inscribed
in the triangle whose angular points are (1, 2), (2, 3), and (3, 1) are
8 + v/lQ and ^^-^^^Q
g ana g .
Find also the coordinates of the centres of the escribed circles.
9. Find the coordinates of the centres, and the radii, of the four
circles which touch the sides of the triangle the coordinates of whose
angular points are the points (6, 0), (0, 6), and (7, 7).
10. Find the position of the centre of the circle circumscribing
the triangle whose vertices are the points (2, 3), (3, 4), and (6, 8).
Find the area of the triangle formed by the straight lines whose
equations are
11. y = x, y = 2x, and y = Sx + 4.
12. y + x=0, y=x + G, and y = 7x + 5.
13. 2y + x-5 = 0, y + 2x-7 = 0, and x-y + l = 0.
14. Sx-'iy + 4a = 0, 2x-By + 4a = 0, and 5x-y + a = 0, proving also
that the feet of the perpendiculars from the origin upon them are
coUinear.
15. y = ax-bc, y = bx-ca, and y = cx-db.
16. y - m.x-\ - , y = in^-\ , and y = moX + - - .
17. y=m^x + Ci, y - m^x + c^, and the axis oiy.
18. y=v\x + c-^, y =m^ + c^, and y=m^ + c^.
19. Prove that the area of the triangle formed by the three straight
lines a^x + h^y + c^ = 0, a^x + K^y + Cg = 0., and a^x + Z>3i/ + Cg = 0 is
80 COORDINATE GEOMETRY. [ExS. X.]
20. Prove that the area of the triangle formed by the three straight
lines
X GOB a + y sin a- Pi = 0, xcos^ + y sin/S-^g^^?
and a; cos 7+?/ sin 7 -2)3 = 0,
^ sin (7 - /3) sin (a - 7) sin {^-a)
21. Prove that the area of the parallelogram contained by the
lines
4y-Sx-a=0, Sy -4:X + a = 0, 4:y-Sx-da=0,
and 3y-4:x + 2a=0 is faK
22. Prove that the area of the parallelogram whose sides are the
straight lines
aiX + biy + Ci = 0, ajX + bjy + dj^ = 0, a^x + b^y + c^^O,
and a^ + b2y + d2=0
IS
«1&2 - ^2^1
23. The vertices of a quadrilateral, taken in order, are the points
(0, 0), (4, 0), (6, 7), and (0, 3) ; find the coordinates of the point of
intersection of the two lines joining the middle points of opposite
sides.
24. The lines a; + 2/ + 1=0, x-y + 2=0, 4x + 2y + S=0, and
x + 2y-4: = 0
are the equations to the sides of a quadrilateral taken in order ; find
the equations to its three diagonals and the equation to the line on
which their middle points lie.
25. Shew that the orthocentre of the triangle formed by the three
straight lines
a a - a
y=mr.x-\ - , y=.m^x-\ - , and y=nux-\ -
is the point
26. -4 and B are two fixed points whose coordinates are (8, 2) and
(5, 1) respectively ; ABP is an equilateral triangle on the side of AB
remote from the origin. Find the coordinates of P and the ortho-
centre of the triangle ABP.
102. £jX. The base of a triangle is fixed ; find the
locus of the vertex when one base angle is double of the
other.
THE STRAIGHT LINE. PROBLEMS.
Let AB be the fixed base of the triangle j take its
middle point 0 as origin, the direc-
tion of 0£ as the axis of x and a
perpendicular line as the axis of y.
Let AO=OB^a.
If P be one position of the O B
vertex, the condition of the problem then gives
lPBA^2^.PAB,
i.e. TT - cf> = 20,
i.e. - tan <^=:tan W (1).
Let P be the point (li, k). "We then have
= tan 0 and -; = tan ^.
h + a h - a
Substituting these values in (1), we have
1{h + a)k
k
h + h
a
(h+ay-k^'
i.e. -{h + aY + ^=2{h^-a%
i.e. k''-3h^-2ah + a^^0.
But this is the condition that the point (A, k) should lie
on the curve
y^-3x'-2ax + a'' = 0.
This is therefore the equation to the required locus.
103. Ex. From a point P perpendiculars PM and
PN are draivn upon two fixed lines which are inclined at an
angle w and meet in a fixed point 0 ; if P move on a fixed
straight line, find the locus of the middle point of MN.
Let the two fixed lines be taken as the axes. Let the
coordinates of P, any position of the
moving point, be {h, Tc).
Let the equation of the straight
line on which P lies be
Ax + By + (7 = 0,
so that we have
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Ah + Bk + C = 0 (1).
L.
82 COORDINATE GEOMETRY.
Draw PL and PL' parallel to the axes.
We then have
0M= OL + LM = OL + LP cos oi^h + Jc cos o>,
and 0J\^= OL' + L'N ^LP + L'P cosoi = k + h cos w.
M is therefore the point {h + k cos co, 0) and J^ is the point
(0, k + h cos co).
Hence, if {x, y) be the coordinates of the middle point
of JfiV, we have
2cc' =h-\-h cos CO (2),
and 2y - k + h cos co (3).
Equations (1), (2), and (3) express analytically all the
relations which hold between x, y\ A, and k.
Also li and k are the quantities which by their variation
cause Q to take up different positions. If therefore between
(1), (2), and (3) we eliminate h and k we shall obtain a
relation between x and y which is true for all values' of h
and k^ i. e. a relation which is true whatever be the position
that P takes on the given straight line.
From (2) and (3), by solving, we have
k ^ 2(a3^-y^cosco) ^^^ ^ ^ 2 {y' - x' cos co) ^
sin^ 00 sin^ co
Substituting these values in (1), we obtain
2 A {x' - y' cos co) + 2^ {y - x cos co) + C sin^ co == 0.
But this is the condition that the point (cc', y) shall
always lie on the straight line
2A{x-y cos oi) + '2£ {y - X cos co) + C sin^ co - 0,
i. e. on the straight line
x{A - £ cos (j>) + y {£ - A cos co) + J C sin- co = 0,
which is therefore the equation to the locus of Q.
104. Ex. A straight line is drawn jjarallel to the
base of a given triangle and its extremities are joined trans-
versely to those of the base; find the locus of the 2>oint
of intersection of the joining lines.
THE STRAIGHT LINE. PROBLEMS.
Let the triangle be OAB and take 0 as the origin and
the directions of OA and OB
as the axes of x and y.
Let OA = a and OB = 6,
so that a and h are given
quantities.
Let A'B' be the straight
line which is parallel to the
base ABj so that
OA ~~0B
and hence OA' = \a and OB' - Xb.
For different values of X we therefore have different
positions of A'B'.
The equation to AB' is
X (say),
and that to A'B is
a Xb
Since F is the intersection of AB' and A'B its coordi-
nates satisfy both (1) and (2). Whatever equation we
derive from them must therefore denote a locus going
through P. Also if we derive from (1) and (2) an equation
which does not contain X, it must represent a locus which
passes through P whatever be the value of A.; in other
words it must go through all the different positions of the
point P.
Subtracting (2) from (1), we have
l(}-l
^%(Ui
b\X
0,
X
a b'
This then is the equation to the locus of P.
always lies on the straight line
b
Hence P
6 - 2
84 COORDINATE GEOMETRY.
which is the straight line OQ where OAQB is a parallelo-
gram.
Aliter. By solving the equations (1) and (2) we
easily see that they meet at the point
\ri"' r?i*^-
Hence, if P be the point (A, k), we have
h = - - L a and k = t zr ^•
Hence for all values of A., i.e. for all positions of the
straight line A'B\ we have
Ji_k
a h '
But this is the condition that the point {h, A), i.e. P,
should lie on the straight line
a h '
The straight line is therefore the required locus.
105. Ex. A variable straight line is drawn through
a given point 0 to cut two fixed straight lines in R and S ;
on it is taken a j)oint P such that
OP''OR'^OS''
shew that the locus of P is a third fixed straight livie.
Take any two fixed straight lines, at right angles and
passing through 0, as the axes and let the equation to the
two given fixed straight lines be
Ax + By + C=0,
and A'x + B'y + G' = 0.
Transforming to polar coordinates these equations are
1 AcosO + B^ine , I A' cos 6 + B' sin 6
THE STRAIGHT LINE. PROBLEMS. 85
1 1
If the angle XOE be 6 the values of ^r^ and ;=r-5 are
therefore
^ cos ^ + ^ sin 0 A' cos $ + B' sin 6
We therefore have
2 _ ^cos^ + .gsin^ A' cos 0 + B' sin 6
0P~~ C C'
The equation to the locus of P is therefore, on again
transforming to Cartesian coordinates,
and this is a fixed straight line.
EXAMPLES. XL
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The base BG (=:2a) of a triangle ABC is fixed; the axes being
BC and a perpendicular to it through its middle point, find the locus
of the vertex A^ when
1. the difference of the base angles is given ( = a).
2. the product of the tangents of the base angles is given ( = X).
3. the tangent of one base angle is m times the tangent of the
other.
4. m times the square of one side added to n times the square of
the other side is equal to a constant quantity c^.
From a point P perpendiculars TM and PN are drawn upon two
fixed lines which are inclined at an angle w, and which are taken as
the axes of coordinates and meet in 0 ; find the locus of P
5. if Oll^ ON be equal to 2c. 6. if OM- ON be equal to 2d.
7. if PM + PN be equal to 2c. 8. if P^I - P^V be equal to 2c.
9. if MN be equal to 2c.
10. if MN pass through the fixed point (a, b).
11. if MN be parallel to the given line y = mx.
86 COORDINATE GEOMETRY. [Exs.
12. Two fixed points A and B are taken on the axes such that
OA = a and OB = h; two variable points A' and B' are taken on the
same axes; find the locus of the intersection of AB' and A'B
(1) when OA' + OB' = OA + OB,
and (2) when __,__=A__.
13. Through a fixed point P are drawn any two straight lines to
cut one fixed straight Hne OX in A and B and another fixed straight
line OY in. C and D ; prove that the locus of the intersection of the
straight lines AG and BD is a straight line passing through 0.
14. OX and OY are two straight lines at right angles to one
another; on OF is taken a fixed point A and on OX any point B;
on AB an equilateral triangle is described, its vertex C being on the
side of AB away from 0. Shew that the locus of 0 is a straight
line.
15. If a straight line pass through a fixed point, find the locus of
the middle point of the portion of it which is intercepted between two
given straight lines.
16. A and B are two fixed points; if PA and PB intersect a
constant distance 2c from a given straight line, find the locus of P.
17. Through a fixed point 0 are drawn two straight lines at right
angles to meet two fixed straight lines, which are also at right angles,
in the points P and Q. Shew that the locus of the foot of the
perpendicular from 0 on PQ is a straight line.
18. Find the locus of a point at which two given portions of the
same straight line subtend equal angles.
19. Find the locus of a point which moves so that the difference
of its" distances from two fixed straight lines at right angles is equal
to its distance from a fixed straight line.
20. A straight line AB, whose length is c, slides between two
given oblique axes which meet at 0 ; find the locus of the orthocentre
of the triangle OAB.
21. Having given the bases and the sum of the areas of a number
of triangles which have a common vertex, shew that the locus of this
vertex is a straight line.
22. Through a given point 0 a straight line is drawn to cut two
given straight lines in R and S ; find the locus of a point P on this
variable straight line, which is such that
(1) 20P.= 0R+0S,
and (2) 0P^=0R.0S.
XI.] THE STRAIGHT LINE. EXAMPLES. 87
23. Given n straight lines and a fixed point 0; through 0 is
drawn a straight line meeting these lines in the points K^, R^, R.^,
...Bm and on it is taken a point R such that
n 1 1 1 1
OR OR^ OR^ OR^ OR^'
shew that the locus of i2 is a straight line.
24. A variable straight line cuts off from n given concurrent
straight lines intercepts the sum of the reciprocals of which is con-
stant. Shew that it always passes through a fixed point.
25. If a triangle ABC remain always similar to a given triangle,
and if the point A be fixed and the point B always move along a
given straight line, find the locus of the point C.
26. -A. right-angled triangle ABC, having C a right angle, is of
given magnitude, and the angular points A and B slide along two
given perpendicular axes; shew that the locus of G is the pair of
straight lines whose equations are y = ±-x.
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27. Two given straight lines meet in 0, and through a given point
P is drawn a straight line to meet them in Q and R; if the
parallelogram OQSR be completed find the equation to the locus
of R.
28. Through a given point 0 is drawn a straight line to meet two
given parallel straight lines in P and Q ; through P and Q are drawn
straight lines in given directions to meet in R ; prove that the locus of
R isa, straight line.
CHAPTER VI.
ON EQUATIONS REPRESENTING TWO OR MORE
STRAIGHT LINES.
106. Suppose we have to trace the locus represented
by the equation
2/2_3a;2/ + 2a;2 = 0 (1).
This equation is equivalent to
{y-x){y-2x) = 0 (2).
It is satisfied by the coordinates of all points which
make the first of these brackets equal to zero, and also by
the coordinates of all points which make the second
bracket zero, i.e. by all the points which satisfy the
equation
and also by the points which satisfy
y-2x = 0 (4).
But, by Art. 47, the equation (3) represents a straight
line passing through the origin, and so also does equa-
tion (4).
Hence equation (1) represents the two straight lines
which pass through the origin, and are inclined at angles of
45° and tan~^ 2 respectively to the axis of x.
107. Ex. 1. Trace the locus xy = 0. This equation
is satisfied by all the points which satisfy the equation
x - O and by all the points which satisfy y - 0, i. e. by
all the points which lie either on the axis of y or on the
axis of X.
EQUATIONS EEPRESENTING STRAIGHT LINES. 89
The required locus is therefore the two axes of coordi-
nates.
Elx. 2. Trace the locus x^ - hx+ &=^ 0. This equation
is equivalent to {x - 2) (a; - 3) = 0. It is therefore satisfied
by all points which satisfy the equation a; - 2 = 0 and also
by all the points which satisfy the equation a? - 3 = 0.
But these equations represent two straight lines which
are parallel to the axis of y and are at distances 2 and 3
respectively from the origin (Art. 46).
Ex. 3. Trace the locus xy - ix - by ■¥ 'lO = 0. This
equation is equivalent to {x - 5) (2/ - 4) = 0, and therefore
represents a straight line parallel to the axis of y at a
distance 5 and also a straight line parallel to the axis of x
at a distance 4.
108. Let us consider the general equation
ax^ + 2hxy + hy'^ ^ 0 (1).
On multiplying it by a it may be written in the form
(«2x.2 + 2ahxy + hhf) - (A^ - ah) y'^ = 0,
L e. \{ax + hy) + y Jh^ - ab\ {(ax + hy) - y Jh^ - ab\ = 0.
As in the last article the equation (1) therefore repre-
sents the two straight lines whose equations are
ax + hy + y Jh^ - ab = 0 (2),
and ax + hy - y J li^ - ab = 0 (3),
each of which passes through the origin.
For (1) is satisfied by all the points which satisfj'- (2),
and also by all the points which satisfy (3),
These two straight lines are real and different if h^>ab,
real and coincident if h^ = ah, and imaginary if h^<ab.
[For in the latter case the coefiicient of y in each of the
equations (2) and (3) is partly real and partly imaginary.]
In the case when h^ < ab, the straight lines, though
themselves imaginary, intersect in a real point. For the
origin lies on the locus given by (1), since the equation (1)
is always satisfied by the values x = 0 and y = 0.
90 COORDINATE GEOMETRY.
109. An equation such as (1) of the previous article,
which is such that in each term the sum of the indices of x
and y is the same, is called a homogeneous equation. This
equation (1) is of the second degree; for in the first term
the index of cc is 2 ; in the second term the index of both x
and 2/ is 1 and hence their sum is 2 ; whilst in the third
term the index of y is 2.
Similarly the expression
3a^ + ix^y - 5xy'^ + 9y^
is a homogeneous expression of the third degree.
The expression
Scc^ + 4:x'^y - 5xy^ + 9y^ - 7xy
is not however homogeneous ; for in the first four terms
the sum of the indices is 3 in each case, whilst in the last
term this sum is 2.
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From Art. 108 it follows that a homogeneous equation
of the second degree represents two straight lines, real and
different, coincident, or imaginary.
110. The axes being rectangular, to jind the angle
between the straight lines given by the equation
aa? + 'ihxy + by- = 0 (1).
Let the separate equations to the two lines be
y - miX = 0 and y~ni^x - 0 (2),
so that (1) must be equivalent to
b (y - mjx) (y - tUox) = 0 (3).
Equating the coefficients of xy and x^ in (1) and (3), we
have
- b {nil + ^2) = 2/i, and bm^n^ = a,
xu i. 2h ^ a
so that m,. + m.y = - - , ana m^m^ - -j .
b b
CONDITIONS OF PERPENDICULARITY. 91
If 0 be the angle between the straight lines (2) we
have, by Art. 66,
1 + m^m^ 1 + m-{nio,
J
F~T 2Vh2-ab
1 0^ a + b
Hence the required angle is found.
111. Condition that the straight lines of the 'previous
article may he iV) 'perpendicular, and (2) coincident.
(1) If« + 5=:0 the value of tan ^ is cx) and hence 0 is
90° j the straight lines are therefore perpendicular.
Hence two straight lines, represented by one equation,
are at right angles if the algebraic sum of the coefficients of
^ and 2/^ be zero.
For example, the equations
x^ - if = 0 and Qx^ + llxy - 6^- - 0
both represent pairs of straight lines at right angles.
Similarly, whatever be the value of h, the equation
a? + Ihx'y - y^ = 0,
represents a pair of straight lines at right angles.
(2) If h^ = ah, the value of tan 0 is zero and hence 6 is
zero. The angle between the straight lines is therefore
zero and, since they both pass through the origin, they are
therefore coincident.
This may be seen directly from the original equation.
For if A^ = ah, i.e. h = J ah, it may be written
ax^ 4- '^Jah x'y + hy- =^0,
i.e. [Jax-\- JhyY=0.
which is two coincident straight lines.
COORDINATE GEOMETRY.
112. Tojind the equation to the straight lines bisecting
the angle between the straight lines given by
ax'' + 2hxy + by^ = 0 (1).
Let the equation (1) represent the two straight lines
L^OM^ and LJ)M.^ inclined at angles B^ and 0.^ to the axis
of £c, so that (1) is equivalent to
b{y -X tan 6-^ (y - x tan 0^ = 0.
Hence
tan 6^ + tan ^o = - ;- , and tan 0^ tan 6^ - -. . .(2).
6 1 ^ b '
Let OA and OB be the required bisectors.
Since z AOL^ = / L^OA,
:. iA0X-e^ = 6^~- lAOX.
:. 2 lAOX^-B^ + e,^.
Also z ^(9X = 90° + z .4 OX.
:. 2 z^OX=180° + ^i + ^2.
Hence, if 6 stand for either of the angles AOX or BOX,
we have
n/i J. /A A\ t^i^ ^1 + ■tan ^2 2A
' 1 - tan t/j tan t'a 6 - «
by equations (2).
But, if (x, 2/) be the coordinates of any point on either
of the lines OA or OB, we have
tan^=^.
X
TWO STRAIGHT LINES. 93
2h , ^^ 2tan^
tan 2^
a 1 - tan^ 0
a; 2xy
x2 - y2 xy
a~b h
This, being a relation holding between the coordinates
of any point on either of the bisectors, is, by Art. 42, the
equation to the bisectors.
113. The foregoing equation may also be obtained in the follow-
ing manner :
Let the given equation represent the straight hnes
y-m^x=0 and y-ni^x^O (1),'
so that iJii + m2= - Y' ^-nd m^m^^- (2).
The equations to the bisectors of the angles between the straight
lines (1) are, by Art. 84,
y-ntj^x _ y-m^x _ y-vi^x _ y-m^x
or, expressed in one equation,
j y-vi^x _ y-jn^x \ J y-m^x y-iihx \_^
{y-m^xf {y-m^-_
1 + m-^ l + ma^
i.e. (1 + m^) {y^ - 2m^xy + m^x^) - (1 + m-^) {y^ - ^m^xy + m^x'^) = 0,
i. e. (m^2 - m^^) {pt^ -y^) + 2 (m^Tng - 1) (Wj - m^) xy = 0,
i. e. (mj + W2) {x^ -y^) + 2 (wij W2 - 1) ici/ = 0.
Hence, by (2), the required equation is
{x^-y-') + 2{~-ljxy = 0,
x^-y"^ _ xy
a-o h
94 COORDINATE GEOMETRY.
EXAMPLES. XII.
Find what straight lines are represented by the following equations
and determine the angles between them.
1. x^-7xy + 12tf=0. 2. 4a;2-24'cz/+ll?/2 = 0.
3. SSx^--71xij-Uif = 0. 4. x^-&x^ + llx-Q = 0.
5. 2/2-16 = 0. 6. tf-xif-Ux^y + 24:X^==0.
7. x^ + 2xysecd + y'^=0. 8. x'^ + 2xy cotd + 7f = 0.
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9. Find the equations of the straight lines bisecting the angles
between the pairs of straight lines given in examples 2, 3, 8, and 9.
10. Shew that the two straight lines
x^ (tan2 0 + cos2 ^j _ 2xy tan d + y^ sin^ ^ = 0
make with the axis of x angles such that the difference of their
tangents is 2.
11. Prove that the two straight lines
{x^ + y^) (cos^ d sin2 a + sin^ 6) =■■ {x t&n a - y sin 6)-
include an angle 2a.
12. Prove that the two straight lines
x^ sia^acoa^9 + 4iXy sin a sin d + y'^[4: cos a - (1 + cos a)^ cos2^] = 0
meet at an angle a.
GENERAL EQUATION OF THE SECOND DEGREE.
114. The most general expression, which contains
terms involving x and y in a degree not higher than the
second, must contain terms involving x^, xy, 7/~, x, y, and a
constant.
The notation vv^hich is in general use for this ex-
pression is
aa^ + 2hxy + hy"^ + 2gx + yy + c (1).
The quantity (1) is known as the general expression of
the second degree, and when equated to zero is called the
general equation of the second degree.
The student may better remember the seemingly
arbitrary coefficients of the terms in the expression (1)
if the reason for their use be given.
GENERAL EQUATION TO TWO STRAIGHT LINES. 95
The most general expression involving terms only of
the second degree in x, y, and z is
aa^ + h'lf' + Gz^ + %fyz + '^gzx + Vixy (2),
where the coefficients occur in the order of the alphabet.
If in this expression we put ;:; equal to unity we get
ao^ -\-hy^ + G + %fy + ^gx + 2hxy,
which, after rearrangement, is the same as (1).
Now in Solid Geometry we use three coordinates x, y,
and z. Also many formulae in Plane Geometry are derived
from those of Solid Geometry by putting z equal to unity.
We therefore, in Plane Geometry, use that notation
corresponding to which we have the standard notation in
Solid Geometry.
115. In general, as will be shewn in Chapter XV.,
the general equation represents a Curve-Locus.
If a certain condition holds between the coefficients of
its terms it will, however, represent a pair of straight lines.
This condition we shall determine in the following
article.
116. To find the condition that the general equation
of the second degree
ax^ + 2hxy + hy'^ + Igx + 2/^/ + c == 0 (1)
Tnay rejyresent tivo straight lines.
If we can break the left-hand members of (1) into two
factors, each of the first degree, then, as in Art. 108, it
will represent two straight lines.
If a be not zero, multiply equation (1) by « and arrange
in powers oi x; it then becomes
a^a? + 2ax {hy + g) = - ahy^ - 2afy - ac.
On completing the square on the left hand we have
«V ^ 2ax {hy + g) + (hy + gf = y- {li? - ah)
■\-2y{gh-af)^g''~-ac,
i.e.
{ax-\-hy+g)=^Jy''{h''-ab) + 2y{gh- of) + g^'-ac ...{2).
96 COORDINATE GEOMETRY.
From (2) we cannot obtain x in terms of y, involving
only terms of the Jirst degree, unless the quantity under the
radical sign be a perfect square.
The condition for this is
i. e. gVi^ - 2qfgh + o?f'^ = g^li? - ahg"^ - acli? + a^bc.
Cancelling and dividing by a, we have the required
condition, viz.
abc + 2fgh - af 2 - bg2 - ch2 = O (3).
117. The foregoing condition may be otherwise obtained thus :
The given equation, multiplied by (a), is
a?x'^-{-2ahxy + aby'^ + 2agx^-2afij + ac = 0 (4).
The terms of the second degree in this equation break up, as in
Art. 108, into the factors
ax + liy-y /Jh^-ab and ax + hy+y fjh^-ab.
If then (4) break into factors it must be equivalent to
{ax + {h-^JhF^ab)y + A}{ax + {h+s/h^-ab)y + B} = 0,
where A and B are given by the relations
a{A+B) = 2ga (5),
A{h+ ^W^^) +B{h- ^h^ - ab) = 2fa (6),
and AB = ac (7).
The equations (5) and (6) give
, , ^ 2fa-2gh
A+B = 2g, and A - B=^^- - =A=..
The relation (7) then gives
Aac = 4:AB = {A+B)^-{A-By-
i.e. {fa-ghf = {g-~ac){h^-ab),
which, as before, reduces to
abc + 2fgh - ap - bg"^ - cli^=^^.
Ex. If a be zero, prove that the general equation will represent
two straight lines if
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If both a and 6 be zero, prove that the condition is 2fg - ch~0.
GENERAL EQUATION TO TWO STRAIGHT LINES. 97
118. The relation (3) of Art. 116 is equivalent to the
expression
<x, h, g
This may be easily verified by writing down the value
of the determinant by the rule of Art. 5.
A geometrical meaning to this form of the relation (3)
will be given in a later chapter.
The quantity on the left-hand side of equation (3) is
called the Discriminant of the General Equation.
The general equation therefore represents two straight
lines if its discriminant be zero.
119. Ex. 1. Prove that the equation
12a;2 + Ixy - lOy"^ + 13a; + 45?/ - 35 = 0
represents tioo straight lines, and find the angle between them.
Here
a=12, /i=|, &=-10, g=^, /=-*/, and c=- 35.
Hence abc + 2fgh - ap - bg^ - ch^
= 12 X ( - 10) X ( - 35) + 2 X ¥- x V- x ^ - 12 x (V)^ - ( - 10) x (V)^
-(-35)(i)2
= 4200 + ^V- - 6075 + i^-^ + i J^5-
The equation therefore represents two straight lines.
Solving it for x, we have
o,,7y + 13 f7y + lBY_10y^-4.mj + S5 fly + isy
_/2%--43\2
~l 24 J •
. , ,7y + 13_ 23y-43
o 4
The given equation therefore represents the two. straight lines
3x=2y-7 and 4x= -5y + 5.
L. r
98 COORDINATE GEOMETRY.
The "m's" of these two lines are therefore f and -f , and the
angle between them, by Art. 66,
Ex. 2. Find the value of h so that the equation
6a;2 + 2hxy + 12y^ + 22a; + 31?/ + 20 = 0
may represent two straight lines.
Here
a = 6, i = 12, ^ = 11, /=V-. andc = 20.
The condition (3) of Art. 116 then gives
20;i2- 341/1 +H*'-^=o,
i.e. (/i -V^) (20/1- 171) = 0.
Hence 7i= V- or ^.
Taking the first of these values, the given equation becomes
6^2 + nxy + 12y^ + 22x + 31?/ + 20 = 0,
Taking the second value, the equation is
20a;2 + 57a;?/ + 40?/2 + ^fa a; + H-2/ + -t- = 0,
EXAMPLES. XIII.
Prove that the following equations represent two straight lines ;
find also their point of intersection and the angle between them.
1. 6?/2-rci/-a;2 + 30?/ + 36 = 0. 2. a;2 - 5;r?/ + 4?/2 + a; + 2?/ - 2 = 0.
3. 3?/2-8a;?/-3a;2-29a; + 3?/-18 = 0.
4. y^ + ^y - 2a;2 - 5x-y -2 = 0.
5. Prove that the equation
a;2 + 6a;?/ + 9?/2 + 4a; + 12?/ - 5 = 0
represents two parallel lines.
Find the value of k so that the following equations may represent
pairs of straight lines :
6. 6a;2+lla;?/-10?/2 + a; + 31?/ + ^ = 0.
7. 12a;2-10a;?/ + 2?/2 + lla;-5?/ + A; = 0.
8. 12a;2 + A;a;?/ + 2?/2 + lla;-5?/ + 2:=0.
9. 6a;2 + a;?/ + A-?/2-ll.r + 43?/-35 = 0.
[EXS. XIII.] EXAMPLES. 99
10. kxy-8x + 9y-12 = 0.
11. x'^ + is^-xij + 'if-5x-7y + k=:0.
12. 12x^ + xy-Qif-2dx + 8y + k = 0.
13. 2x^ + oiyy-y'^ + kx + 6y-d = 0.
14. x^ + kxy + y^-5x-ly + Q = 0.
15. Prove that the equations to the straight lines passing through
the origin which make an angle a with the straight line y + x = 0 are
given by the equation
x^ + 2xy sec 2a + y^ = Q.
16. What relations must hold between the coordinates of the
equations
(i) ax^ + hy'^ ■^cx-\-cy = 0,
and (ii) ay'^ + lxy + dy -{-ex=zQ^
so that each of them may represent a pair of straight lines ?
17. The equations to a pair of opposite sides of a parallelogram
are
a;2-7a; + 6 = 0 and 2/^- 14?/ +40 = 0 ;
find the equations to its diagonals.
120. To prove that a homogeneous equation of the nth
degree represents n straight li7ies, real or imaginary, which
all pass through the orighi.
Let the equation be
y'^ + A^xy''-^ + A^xY'-^ + A.^x'y''-^ 4- . . . + A.^x'' =-- 0.
On division by x^, it may be written
This is an equation of the nth degree in - , and hence
JO
must have n roots.
Let these roots be -m^, m^, mg, ... m„. Then (C. Smith's
Algebra, Art. 89) the equation (1) must be equivalent to
the equation
t
The equation (2) is satisfied by all the points which
satisfy the separate equations
tLf *^ tKf
7 - 2
100 COORDINATE GEOMETRY.
i.e. by all the points which lie on the n straight lines
y - nijX = 0, y - m^x = 0, ... y - in^x = 0,
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all of which pass through the origin. Conversely, the
coordinates of all the points which satisfy these n equa-
tions satisfy equation (1). Hence the proposition.
121. Ex. 1. The equation
which is equivalent to
{y-x){y-'lx){y-^x) = Q,
represents the three straight lines
y-x=.^, ?/-2a: = 0, and y-3a; = 0,
all of which pass through the origin.
Ex. 2. The equation y^ - 5y^ + 6y = 0,
i.e. y{y-2){y-S) = 0,
similarly represents the three straight lines
y=0, y = 2, and y = S,
all of which are parallel to the axis of x.
122. To find the equation to the two straight lines
joining the origin to the points in which the straight line
Ix + my = n (1)
meets the locus whose equation is
ax^ + 2hxy + hy"^ + "Igx + 2fy + c = 0 (2).
The equation (1) may be written
Ix + my
n
The coordinates of the points in which the straight line
meets the locus satisfy both equation (2) and equation (3),
and hence satisfy the equation
ax' + Vum) + by^ + 2 {gx +/y) '-^^ + c ('-^^)' = 0
[For at the points where (3) and (4) are true it is clear
that (2) is true.]
Hence (4) represents so7ne locus which passes through
the intersections of (2) and (3).
STRAIGHT LINES THROUGH THE ORIGIN. 101
But, since the equation (4) is homogeneous and of the
second degree, it represents two straight lines passing
through the origin (Art. 108).
It therefore must represent the two straight lines join-
ing the origin to the intersections of (2) and (3).
123. The preceding article may be illustrated geo-
metrically if we assume that the equation (2) represents
some such curve as PQRS in the figure.
Let the given straight line cut the curve in the points
P and Q.
The equation (2) holds for all points on the curve PQRS.
The equation (3) holds for all points on the line PQ.
Both equations are therefore true at the points of
intersection P and Q.
The equation (4), which is derived from (2) and (3),
holds therefore at P and Q.
But the equation (4) represents two straight lines, each
of which passes through the point 0.
It must therefore represent the two straight lines OP
and OQ.
124. Ex. Prove that the straight lines joining the origin to the
points of intersection of the straight line x-y=:2 and the curve
5a;2 + 12a;i/ - 82/2 + 8a; - 4y + 12 = 0
make equal angles with the axes.
As in Art. 122 the equation to the required straight lines is
5a;2 + 12a;?/-8i/2+(8a;-4i/)^^ + 12fc^Y = 0 (1),
102 • COORDINATE GEOMETRY.
For this equation is homogeneous and therefore represents two
straight lines through the origin; also it is satisfied at the points
where the two given equations are satisfied.
Now (1) is, on reduction,
so that the equations to the two lines are
y = 2x and y= -2x.
These lines are equally inclined to the axes.
125. It was stated in Art. 115 that, in general^ an
equation of the second degree represents a curve- line,
including (Art. 116) as a particular case two straight lines.
In some cases however it will be found that such
equations only represent isolated points. Some examples
are appended.
EjX. 1. What is represented hy the locus
{x-y + cf+{x + y-cY = 01 (1).
We know that the sum of the squares of two real
quantities cannot be zero unless each of the squares is
separately zero.
The only real points that satisfy the equation (1)
therefore satisfy both of the equations
cc - 2/ + c = 0 and x + y - G = 0.
But the only solution of these two equations is
re = Oj and y = c.
The only real point represented by equation (1) is therefore
The same result may be obtained in a different manner.
The equation (1) gives
{x-y-VGY^-(x + y-cf,
It therefore represents the two imaginary straight lines
and X (1 + J^l)-y (I - J^) + c(l - J~l) = 0.
EQUATIONS REPRESENTING ISOLATED POINTS. 103
Each of these two straight lines passes through the
real point (0, c). We may therefore say that (1) represents
two imaginary straight lines passing through the point
£jX. 2. What is represented hy the equation
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As in the last example, the only real points on the locus
are those that satisfy both of the equations
oc^ - a^ - 0 and y^ - h^ = 0,
i.e. x = =i= a, and y = d=h.
The points represented are therefore
Ex. 3. WTiat is represented by the equation
The only real points on the locus are those that satisfy
all three of the equations
x = 0, 2/=0, and a = 0.
Hence, unless a vanishes, there are no such points, and
the given equation represents nothing real.
The equation may be written
so that it represents points whose distance from the origin
is asl - \. It therefore represents the imaginary circle
whose radius is asJ - 1 and whose centre is the origin.
126. Es. 1. Obtain the condition that one of the straight lines
given hy the equation
ax^-\-2hxy + by^ = 0 (1)
may coincide with one of those given by the equation
a'x^ + 2h'xy + bY=0 (2).
Let the equation to the common straight line be
y-m^x = 0 (3).
The quantity y -m^x must therefore be a factor of the left-hand of
both (1) and (2), and therefore the value y = m^x must satisfy both (1)
and (2).
104 COORDINATE GEOMETRY.
"We therefore have
bmi^ + 2hm^ + a=:0 (4),
and b'm^^ + 2h'm^ + a' = 0 (5).
Solving (4) and (5), we have
2 {ha' - h'a) ~ aV -a'b~2 {bh' - b'h) '
^ ha'-h'a_ ^_ f ab'-a'b ]^
so that we must have
{ab' - a'&)2 = 4 {ha' - h'a) {bh' - b'h) .
Ex. 2. Prove that the equation
m{x^-Sxy^) + y^-3x^y=0
represents three straight lines equally inclined to one another.
Transforming to polar coordinates (Art. 35) the equation gives
m {GQS^d- 3 cos d sin2^) + sin3^ - 3 cos^^ sin ^ = 0,
i.e. m(l-3tan2^) + tan3^-3tan^ = 0,
3tan^-tan3^ ^ „^
If m=tan a, this equation gives
tan 3^ = tan a,
the solutions of which are
3^ = a, or 180° + a, or 360° + a,
The locus is therefore three straight lines through the origin
inclined at angles
^, 60° + |, and 120° + ^
to the axis of x.
They are therefore equally inclined to one another.
Ex. 3. Prove that two of the straight lines represented by the
equation
ax'^ + bx^y + cxy^ + dy^ = 0 (1)
will be at right angles if
a^ + ac + hd + d^ = 0.
Let the separate equations to the three lines be
y-miX = 0, y-m2X=0, and y-m^x=0,
EXAMPLES. 105
so that the equation (1) must be equivalent to
c
and therefore mj^+m2+m^= -- (2),
Wom3 + m3?7ij + 77ijm2 = -^ (3)i
and m^m^m^ = - -^ (4).
If the first two of these straight lines be at right angles we have,
in addition,
711^111.2= -1 (5).
From (4) and (5), we have
a
and therefore, from (2), a c + a
The equation (3) then becomes f c + a\ _b
i.e. a^ + ac + bd + d^ = 0.
EXAMPLES. XIV.
1. Prove that the equation
y^-x^ + 3xy (y -x) = 0
represents three straight lines equally inclined to one another.
2. Prove that the equation
y^ (cos a + fJ3 sin a) cos a-xy (sin 2a - ^^3 cos 2a)
+ x^ (sin a - ^3 cos a) sin a = 0
represents two straight lines inclined at 60° to each other.
Prove also that the area of the triangle formed with them by the
straight line
(cosa-/sy3 sin a) 2/ -(sin a + ;^3cos a)a; + a-=0
a2
V3'
and that this triangle is equilateral.
3. Shew that the straight lines
(^2 _ 3^2) ^2 + sABxy + [B'^ - 3A^) y^=0
form with the Hne Ax + By + C = 0 an equilateral triangle whose area
106 COORDINATE GEOMETRY. [ExS.
_ 4. Find the equation to the pair of straight lines joining the
origin to the intersections of the straight line y = mx + c and the curve
Prove that they are at right angles if
2c2 = a2(l + m2).
5. Prove that the straight lines joining the origin to the points
of intersection of the straight line
kx + hy = 2hk
with the curve {^-h)^+{y- k)^ = c^
are at right angles if h^+k^=c^.
6. Prove that the angle between the straight lines joining the
origin to the intersection of the straight line y = 3x + 2 with the curve
x'^ + 2xy + Sy^ + 'ix+8y-ll = 0 istan-i?^.
7. Shew that the straight lines joining the origin to the other two
points of intersection of the curves whose equations are
ax^ + 2hxy + by^ + 2gx=0
and a'x'^ + 2h'xy + hY + ^g'x = 0
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will be at right angles if
What loci are represented by the equations
8. x^-y^=0. 9. x'^-xy = 0. 10. xy-ay = 0.
11. x^-x^-x + l = 0. 12. x^-xy^ = 0. 13. x^ + y^ = 0.
14. x^ + y^=0. 15. x^y = 0. 16. {x'--l){y^-^)=0.
17. {x^-lf + {y^-4y=0. 18. {y-mx-cY + {y-m'x-c')^=0,
19. {a;2-a3)2(^2_52)2 + c'*(?/2-a2) = 0. 20. {x-a)^-y^=0.
21, (x + y)^-c^=0. 22. r=a sec (<?- a).
23. Shew tliat the equation
bx^-2hxy + ay^=0
represents a pair of straight lines which are at right angles to the pair
given by the equation
ax^ + 2 Jixy + by^ = 0.
24. If pairs of straight lines
x^-2pxy -y^=0 and x^-2qxy-y^=0
be such that each pair bisects the angles between the other pair, prove
thatjpgr= -1.
25. Prove that the pair of lines
a^x'^ + 2h{a + b)xy + b^y^=zO
is equally inclined to the pair
ax^ + 2hxy + by^=0.
XIV.] EXAMPLES. 107
26. Shew also that the pair
is equally inclined to the same pair.
27. If one of the straight lines given by the equation
ax^ + Ihxy + hy^ = 0
coincide with one of those given by
a'x^ + 2h'xy + bY=0,
and the other lines represented by them be perpendicular, prove that
ha'b' h'ah , - -
0 -a b-a ^
28. Prove that the equation to the bisectors of the angle between
the straight lines ax^ + 2hxy + by'^=0 is
h {x^ - y^) + {b - a) xy = {ax^ - by^) cos w,
the axes being inclined at an angle w.
29. Prove that the straight lines
ax^-\-1'kxy + by'^=zQ
make equal angles with the axis of a; if 7t = acosw, the axes being
inclined at an angle w.
30. If the axes be inclined at an angle w, shew that the equation
x^ + 2ocy cos w + 2/^cos 2a; = 0
represents a pair of perpendicular straight lines.
31. Shew that the equation
cos 3a {x^ - dxy^) + sin 3a (y^ - Sx'^y) + 3a (x^ + y^) -^a^ = 0
represents three straight lines forming an equilateral triangle.
Prove also that its area is 3 sJSa^.
32. Prove that the general equation
ax^ + 2Jixy + by^ + 2gx + 2fy + c=zO
represents two parallel straight lines if
h^ = ab and bg^=^af^.
Prove also that the distance between them is
V a(a-
ac
33. If the equation
ax'^ + 2hxy + by^ + 2gx + 2fy + c = 0
represent a pair of straight lines, prove that the equation to the third
pair of straight lines passing through the points where these meet the
axis is
ax^ - 2hxy + by^ + 2gx + 2fy + c + -^^xy = 0.
108 COORDINATE GEOMETRY. [Exs. XIV.]
34. If the equation
as(^ + 2hxy + by^ + 2gx + 2fy + c = 0
represent two straight lines, prove that the square of the distance of
their point of intersection from the origin is
35. Shew that the orthocentre of the triangle formed by the
straight lines
ax^ + 2kxy + by^=0 and lx + viy = l
is a point (x\ y') such that
I ~ m~ aw? - 2hlm + bP '
36. Hence find the locus of the orthocentre of a triangle of which
two sides are given in position and whose third side goes through a
fixed point.
37. Shew that the distance between the points of intersection of
the straight Hne
X cos a + y sin a-p = 0
with the straight lines ax^ + 2 hxy + dy^=0
2pjh^-ab
& cos^a - 2/i cos a sin a + a sin^ a '
Deduce the area of the triangle formed by them.
38. Prove that the product of the perpendiculars let fall from the
point {x', y') upon the pair of straight lines
ax^ + 2hxy + by"^ - 0
ax''^ + 21tx'y' + by'^
^^ J{a-bf + 4:h^ '
39. Shew that two of the straight lines represented by the
equation
ay"^ + bocy^ + cx^y^ -}-dx^y + ex'^ = 0
will be at right angles if
40. Prove that two of the lines represented by the equation
ax^ + bx^ y + cx^ y^ + dxy^ + ay^ = 0
will bisect the angles between the other two if
c + Qa=0 and b + d = 0.
41. Prove that one of the lines represented by the equation
ax^ + bx"^ y + cxy^ + dy^ = 0
will bisect the angle between the other two if
{3a + c)^{bc + 2cd - Sad) = (6+ Sd)^{bc + 2ab - 3ad).
CHAPTER yil.
TRANSFORMATION OF COORDINATES.
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127. It is sometimes found desirable in the discussion
of problems to alter the origin and axes of coordinates,
either by altering the origin without alteration of the
direction of the axes, or by altering the directions of the
axes and keeping the origin unchanged, or by altering the
origin and also the directions of the axes. The latter case
is merely a combination of the first two. Either of these
processes is called a transformation of coordinates.
We proceed to establish the fundamental formulae for
such transformation of coordinates.
128. To alter the origin of coordinates without altering
the directions of the axes.
Let OX and 0 Z be the original axes and let the new
axes, parallel to the original, be
O'X' and O'Y'.
Let the coordinates of the new
origin 0\ referred to the original
axes be h and k, so that, if O'L be
perpendicular to OX, we have
OL = h and LO' = h.
Let P be any point in the plane
of the paper, and let its coordinates, referred to the original
axes, be x and y, and referred to the new axes let them be
x' and y'.
Draw PN perpendicular to OX to meet OX' in N'.
Y'
p
N'
X'
N
COORDINATE GEOMETRY.
Then
ON^x, NF = y, 0'N' = x, and N'F^y'.
We therefore have
X ^ 0N= OL + O'N' = h + x\
and y = FP = LO' + N'P = k + y'.
The origin is therefore transferred to the point Qi, k) when
we substitute for the coordinates x and y the quantities
X + h and y' + k.
The above article is true whether the axes be oblique
or rectangular.
129. To change the direction of the axes of coordinates,
without changing the origin, both systems of coordinates being
rectangular.
Let OX and OF be the original system of axes and OX'
and OY' the new system, and let
the angle, XOX' , through which
the axes are turned be called 0. Y'
Take any point F in the plane
of the paper.
Draw FN and FN' perpen-
dicular to OX and OX', and also
N'L and N'M. perpendicular to OX and FN.
If the coordinates of F, referred to the original axes,
be X and y, and, referred to the new axes, be xi and y', we
have
ON^x, NF = y, ON'^x', and N'F = y.
The angle
MFN' - 90° - z MN'F^ l MN'O = z XOX' = 6.
We then have
X = 0N= OL -MN'= ON' cobO-N'F sine
= x' cos 0 - y sin 0 (1),
and y = NF = LN' + MF= ON' sin 6 + N'F cos $
= oj' sin ^ + 2/' cos ^ (2).
CHANGE OF AXES. Ill
If therefore in any equation we wish to turn the axes,
being rectangular, through an angle 0 we must substitute
X' cos ^ - y' sin 0 and x' sin ^ + y' cos 0
for X and y.
"When we have both to change the origin, and also the
direction of the axes, the transformation is clearly obtained
by combining the results of the previous articles.
If the origin is to be transformed to the point (Ji, k)
and the axes to be turned through an angle 6, we have to
substitute
h + X cos 0 - y sin 6 and k + x sin 0 + y cos 6
for X and y respectively.
The student, who is acquainted with the theory of projection of
straight lines, will see that equations (1) and (2) express the fact that
the projections of OP on OX and OY are respectively equal to the
sum of the projections of ON' and N'P on the same two lines.
130. Ex. 1. Transform to parallel axes through the 'point ( - 2, 3)
the equation
2a;2 + 4a;2/ + 5^/2 - 4^ - 22?/ + 7 =: 0.
We substitute x=x' -2 and y=y' + S, and the equation becomes
2 (x' - 2)3 + 4 (a;' - 2) (?/' + 3) + 5 (?/' + 3)2 - 4 (x' - 2) - 22 (i/' + 3) + 7 = 0,
i.e. 2x'^ + 4xy + 5y'^-22 = 0.
Ex. 2. Transform to axes inclined at 30° to the original axes the
equation
x'^ + 2fj3xy-y'^=2a^.
For X and y we have to substitute
flj' cos 30° - ?/ sin 30° and a;' sin 30° + ?/' cos 30°,
The equation then becomes
(a:V3 -2/T + 2 V3 (^V3 -2/') (^' + 2/V3) - (^' + 2/V3)'=8a2,
112 COORDINATE GEOMETRY.
EXAMPLES. XV.
1. Transform to parallel axes through the point (1, -2) the
equations
(1) y'^-4:X + 4y + 8 = 0,
and (2) 2x^ + y^-4:X + 4y = 0.
2. What does the equation
become when it is transferred to parallel axes through
(1) the point {a-c, b),
(2) the point {a, b-c)?
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3. What does the equation
{a-b){x^ + y^-)-2abx=0
become if the origin be moved to the point ( , 0 ) ?
4. Transform to axes inclined at 45° to the original axes the
equations
(2) nx^-l&xy + ny^ = 225,
and (3) y^ + x^ + 6x^y^=2.
5. Transform to axes inclined at an angle a to the original axes
the equations
and (2) a^ + 2xy tan 2a- y^=aK
6. If the axes be turned through an angle tan~i 2, what does the
equation Axy - 3x^ = a^ become ?
7. By transforming to parallel axes through a properly chosen
point {h, k), prove that the equation
12ciP-10xy + 2y^ + llx-5y + 2 = 0-
can be reduced to one containing only terms of the second degree.
8. Find the angle through which the axes may be turned so that
the equation Ax + By + 0=0
may be reduced to the form a; = constant, and determine the value of
this constant.
131. The general proposition, which is given in the
next article, on the transformation from one set of oblique
axes to any other set of oblique axes is of very little
importance and is hardly ever required.
CHANGE OF AXES. 113
■*132. To change from one set of axes, inclined at an
angle w, to another set, inclined at an angle w', the origin
remaining unaltered.
Let OX and (9Fbe the original axes, OX' and OY' the
new axes, and let the angle XOX' be 0.
Take any point P in the plane of the paper.
Draw PN and PN' parallel to 07 and OY' to meet OX
and OX' respectively in N and iV", PL perpendicular to OX,
and N'M and X'M' perpendicular to OL and LP.
Now
z PNL = L YOX = 0), and PN'M' = Y'OX = oi' + B.
Hence if
OX^x, NP^y, ON'^x, s.rvAN'P^y',
we have y s>\n.oi = NP ^irna^ LP = MX' + M'P
= OX' sin 0 + X'P sin (w' + 6),
so that y sin w = cc' sin 6 + y' sin (w' + ^) (1).
Also
x-ryco&oi^OX+XL=^OL=^OM-\-X'M'
= x cosO + y'cos{oy' + e) (2).
Multiplying (2) by sinw, (1) by cosw, and subtracting,
we have
X sin (o = x sin (w - ^) + y' sin (oi - w - 6) (3).
[This equation (3) may also be obtained by drawing a perpen-
dicular from P upon OT and proceeding as for equation (1).]
The equations (1) and (3) give the proper substitutions
for the change of axes in the general case.
As in Art. 130 the equations (1) and (2) may be obtained by
equating the projections of OP and of ON' and N'P on OX and a
straight line perpendicular to OX.
114 COORDINATE GEOMETKY.
'^133. Particular' cases of the preceding article.
(1) Suppose we wish to transfer our axes from a
rectangular pair to one inclined at an angle w'. In this
case (0 is 90°, and the formulse of the preceding article
become
x = x' cos 6 + y' cos (w' + 6),
and y = ^' sin 6 + y sin (w' + 6).
(2) Suppose the transference is to be from oblique
axes, inclined at w, to rectangular axes. In this case co' is
90°, and our formulse become
X sin isi = X sin (w - ^) - y' cos (w - 0)^
and y sin o> = a:;' sin 6 + y' cos ^.
These particular formulse may easily be proved in-
dependently, by drawing the corresponding figures.
Ex. Transform the equation -2 - f^ = l from rectangular axes to
axes inclined at an angle 2a, the new axis of x being inclined at an angle
- a to the old axes and sin a being equal to - , .
Here 6= -a and w' = 2a, so that the formulse of transformation
(1) become
a; = (a;' + y) cos a and y = [y' - x') sin a.
Since sin a = , , we have cos a = , , and hence the
x/a2+&2 Ja'^ + b'^
given equation becomes
i.e. x'y'=i{a'^ + b^).
■*134. The degree of an equation is unchanged hy any
transformation of coordinates.
For the most general form of transformation is found
by combining together Arts. 128 and 132, Hence the
most general formulse of transformation are
, sin (oi - 6) , sin (o> - w' - 6)
sm 0) sm w
and y = k-vx -. - + y ^ .
smo) sinw
CHANGE OF AXES. 115
For X and y we have therefore to substitute expressions
in X and y' of the first degree, so that by this substitution
the degree of the equation cannot be raised.
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Neither can, by this substitution, the degree be lowered.
For, if it could, then, by transforming back again, the
degree would be raised and this we have just shewn to be
impossible.
*135. If by any change of axes, without change of origin, the
quantity ax^ + Ihxy + &i/^ become
a'x'^ + 2h'xY + by%
the axes in each case being rectangular, to prove that
a + b = a' + b', and ab-h^ = a'b' -h'^.
By Art. 129, the new axis of x being inclined at an angle 6 to the
old axis, we have to substitute
a;'cos^-2/'sin^ and x' sin 0 + y' cos. 6
for X and y respectively.
Hence ax"^ + 2hxy + by'^
= a{x'cosd-y' sin ef + 2h {x' cose~y' sin d) {x' sin d + y' cos d)
+ b{x'smd + y'cos6)^
= x'^ [a cos2 e + 2h cos Osind + b sin^ d]
+ 2x'y' [-acosdsind + h (cos^ d - sin^i?) + & cos ^ sin 9]
+ y'^ \a sin2 d-2h cos ^ sin ^ + & cos^ d].
We then have
tt' = a cos^ ^ + 2/i cos ^ sin ^ + 6 sin^ (9
b' = aBin^d- 2h cos ^ sin ^ + 6 cos^^
= i[(a + 6)-(a-&)cos2^-27isin2^] (2),
and /i'= -acos^sin^ + /i(cos2^-sin2^) + &cos^sin^
= i[2;icos2^-(a-&)sin2^j (3).
By adding ( 1) and (2) , we have «' + &'=« + &.
Also, by multiplying them, we have
4a'&' = (a + 6)2 - { (a - h) cos 2d +-2h sin 26}^.
Hence 4a'&' - 47i'2
= (a + 6)2 _ i{2h sin 2d + {a- b) cos 20]^+ {2h cos 20 -{a- b) sin 20}^
= (a + &)2 _ [(a - 6)2 + 4/i2] = 4a6 - 4h%
so that a'b'-h'^ = ab-h^.
136. To find the angle through which the axes must be turned so
that the expression ax^ + 2hxy + by^ may become an expression in which
there is no term involving x'y'.
8 - 2
116 COORDINATE GEOMETRY.
Assuming the work of the previous article the coefficient of x'y'
vanishes if h' be zero, or, from equation (3), if
27icos2^ = (a-Z>)sin2^,
2h
i.e. if tan 2(9= -.
a~h
The required angle is therefore
i tan~i
*137. The proposition of Art. 135 is a particular
case, when the axes are rectangular, of the folloM^ing more
general proposition.
If hy any change of axes, without change of origin, the
quantity ax- + ^hxy + 6?/" becomes a'x'^ + 2h'xy + 6'^/^, then
a + b - 2h cos w a' + b' - 2h' cos w'
sin^ lo sin^ w' '
ab-h^ db'-K''
sm" (i) sm^ o)
0) a7id oi' being the angles between the original and final indrs
of axes.
Let the coordinates of any point P, referred to the
original axes, be x and y and, referred to the final axes, let
them be x and y .
By Art. 20 the square of the distance between P and
the origin is o^ + 2xy cos a> + 2/^, referred to the original axes,
and x'^ + 2xy cos w' + y"^^ referred to the final axes.
We therefore always have
X- + Ixy cos w + 2/^ = a;'^ + ^xy cos m -vy''^ (1).
Also, by supposition, we have
ax"" + 2hxy + by"" = ax'- + 2h'xy' + b'lj^ (2).
Multiplying (1) by \ and adding it to (2), we therefore have
x^ (« + X) + 2xy (h + X cos to) +y^ (b + X)
= x"" (a' + X) + 2x'y' {h' + X cos o>') + y" {b' + X) . . .(3).
If then any value of X makes the left-hand side of (3) a
perfect square, the same value must make the right-hand
side also a perfect square.
But the values of X which make the left-hand a perfect
square are given by the condition
(h + X cos (o)2 -(a + X) {b + X),
EXAMPLES. 117
i.e. by
\2 (1 - cos^ w) + X{a + h-2h cos oi) + ab- h^ = 0,
^„ . a + h - 2hcos(ji ab - h^^ ,,,
In a similar manner the values of A, which make the
right-hand side of (3) a perfect square are given by the
equation
^a' + b'-2h'cosoy' a'b'-h'^ ^
sm^ to sm'^ci) ^ '
Since the values of \ given by equation (4) are the same
as the values of X given by (5), the two equations (4) and
(5) must be the same.
Hence we have
a + b - 2h cos cu a' + b' - 2h' cos co'
and
sin^ (o sin^ co'
ab-h2 a'b'-h'2
sin2 0) sin2 cd'
EXAMPLES. XVI.
1. The equation to a straight line referred to axes inclined at 30°
to one another is y = 2x + l. Find its equation referred to axes
inclined at 45°, the origin and axis of x being unchanged.
2. Transform the equation 2x'^ + 3 sjSxy + Sy^ = 2 from axes
inclined at 30° to rectangular axes, the axis of x remaining
unchanged.
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3. Transform the equation x^ + xy + y^ = 8 from axes inclined at
60° to axes bisecting the angles between the original axes.
4. Transform the equation y'^-\-4cy cot a - 4cc=0 from rectangular
axes to oblique axes meeting at an angle a, the axis of x being kept
the same.
5. If aj and y be the coordinates of a point referred to a system of
obUque axes, and x' and y' be its coordinates referred to another
system of oblique axes with the same origin, and if the formulae of
transformation be
x=mx' + ny' and y - m'x'-\-n'y\
prove that » , ,. ., = - 7- .
CHAPTER VIII.
THE CIRCLE.
138. Def. A circle is the locus of a point which
moves so that its distance from a fixed point, called the
centre, is equal to a given distance,
called the radius of the circle.
The given distance is
139. To find the equation to a circle, the axes of coordi-
nates being two straight lines through its centre at right
angles.
Let 0 be the centre of the circle and let a be its radius.
Let OX and OF be the axes of
coordinates.
Let P be any point on the circum-
ference of the circle, and let its coordi-
nates be X and y.
Draw PJf perpendicular to OX and
join OP.
Then (Euc. I. 47)
OM^ + MP' = a',
i.e. x2 + y2 = a2.
This being the relation which holds between the coordi-
nates of any point on the circumference is, by Art. 42, the
required equation.
140. To find the equation to a circle referred to any
rectangular axes.
THE CIRCLE.
Let OX and (9 F be the two rectangular axes.
Let C be the centre of the
circle and a its radius.
Take any point P on the
circumference and draw per-
pendiculars CM and FN upon
OX ; let F be the point (x, y).
Draw GL perpendicular to
NF.
Let the coordinates of G be
h and h ; these are supposed to be known.
We have GL = MN= ON- OM=x- h,
and LF = NF-NL = NF-MG^y-h.
Hence, since GL^ + LF^ = GF\
we have (x-h)2+ (y-k)2 = a2 (1).
This is the required equation.
Ex. The equation to the circle, whose centre is the point ( - 3, 4)
and whose radius is 7, is
i.e. a;2 + 2/2 + 6a:-8?/ = 24.
141. Some particular cases of the preceding article may be
noticed :
(a) Let the origin 0 be on the circle so that, in this case,
i.e. h^ + Jc'^=a^.
The equation (1) then becomes
i.e. x'^ + y^-2hx-2ky = 0.
(jS) Let the origin be not on the curve, but let the centre lie on
the axis of x. In this case k = 0, and the equation becomes
(7) Let the origin be on the curve and let the axis of re be a
diameter. We now have fe = 0 and a = h, so that the equation becomes
x^ + t/-2hx = 0.
(8) By taking 0 at C, and thus making both h and k Tiero, we
have the case of Art. 139.
120 COORDINATE GEOMETRY.
(e) The circle will touch the axis of x if MG be equal to the
radius, i.e. if k = a.
The equation to a circle touching the axis of x is therefore
x'^ + 7f-2hx-2ky + h'^ = Q.
Similarly, one touching the axis of y is
x^ + y^-2hx-2ky + k^ = 0.
142. To prove that the equation
always represents a circle for all values of g,f and c, and to
find its centre and radius. [The axes are assumed to be
rectangular.]
This equation may be written
i.e. {X + gf 4- {y +ff = {JfTp^cf.
Comparing this with the equation (1) of Art. 140, we
see that the equations are the same if
h = -g, h^-f and a = Jg"" -hf^-c.
Hence (1) represents a circle whose centre is the point
( - ^, - f), and whose radius is Jg^-^f^ - c.
If g^ +f' > c, the radius of this circle is real.
If g^ +f'^ = c, the radius vanishes, i. e. the circle becomes
a point coinciding with the point ( - ^, - f). Such a circle
is called a point-circle.
If g^ +f^ < c, the radius of the circle is imaginary. In
this case the equation does not represent any real geo-
metrical locus. It is better not to say that the circle does
not exist, but to say that it is a circle with a real centre
and an imaginary radius.
Ex. 1. The equation x"^ + ij^ + 4^x - 6y = 0 can be written in the
form
{x + 2f + iy-Bf = lS = {JlS)\
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and therefore represents a circle whose centre is the point ( - 2, 3) and
whose radius is >,yi3.
GENERAL EQUATION TO A CIRCLE. 121
Ex. 2. The equation 45a;2 + 45?/2 - 60x- + SQy + 19 = 0 is equivalent
to
and therefore represents a circle whose centre is the point (f , -|) and
whose radius is - =- .
143. Condition that the general equation of the second
degree may represent a circle.
The equation (1) of the preceding article, multiplied by
any arbitrary constant, is a particular case of the general
equation of the second degree (Art. 114) in which there is
no term containing xy and in which the coefficients of x^
and y^ are equal.
The general equation of the second degree in rectangular
coordinates therefore represents a circle if the coefficients
of x^ and y^ be the same and if the coefficient of xy
be zero.
144. The equation (1) of Art. 142 is called the
general equation of a circle^ since it can, by a proper
choice of g, f and c, be made to represent any circle.
The three constants g, f, and c in the general equation
correspond to the geometrical fact that a circle can be found
to satisfy three independent geometrical conditions and no
more. Thus a circle is determined when three points on it
are given, or when it is required to touch three straight
lines.
145. To find the equation to the circle which is described on the
line joining the points {x-^ , y-^) and (x^ , 2/2) ^^ diameter.
Let A be the point (a;^, y-^) and B be the point {x^, 2/2) » ^^^ ^^^ ^^^
coordinates of any point P on the circle be h and k.
The equation to AP is (Art. 62)
y-yi=j^i^-^i) (1).
and the equation to BP is
But, since APB is a semicircle, the angle APB is a right angle,
and hence the straight Unes (1) and (2) are at right angles.
COOKDINATE GEOMETRY.
Hence, by Art. 69, we have
^-Vi _ h^ljh^ _ 1
i.e. (h-x-^){h-x^) + {k-yj}{Jc-y2) = 0.
But this is the condition that the point {h, k) may lie on the curve
whose equation is
This therefore is the required equation.
146. Intercepts made on the axes by the circle whose equation is
ax^ + ay^ + 2gx + 2fy + c = 0 (1).
The abscissae of the points where the circle (1) meets the axis of x,
i.e. y = 0, are given by the equation
ax^ + 2gx + c = 0 (2).
The roots of this equation being Xj^ and x^ ,
we have
and
(Art. 2.
Hence
V a2
a a
Again, the roots of the equation (2) are both imaginary if g^<ac.
In this case the circle does not meet the axis of x in real points, i.e.
geometrically it does not meet the axis of x at all.
The circle will touch the axis of x if the intercept A-^A^, be just
zero, i.e. ii g^ = ac.
It will meet the axis of x in two points lying on opposite sides of
the origin 0 if the two roots of the equation (2) are of opposite signs,
i. e. if c be negative.
147. Ex.1. Find the equation to the circle which passes through
the points (1, 0), (0, - 6), and (3, 4).
Let the equation to the circle be
x^ + y^ + 2gx + 2fy + c = 0 (1).
Since the three points, whose coordinates are given, satisfy this
equation, we have
36-12/+c=0 (3),
and 25 + 6^ + 8/+c = 0 (4).
EXAMPLES. 123
Subtracting (2) from (3) and (3) from (4), we have
2^ + 12/= 35,
and 65f + 20/=ll.
Hence /=*/- and g=z -iJ-.
Equation (2) then gives c = ^-.
Substituting these values in (1) the required equation is
4a;2 + 4i/2 - 142a; + 47a; + 138 = 0.
Ex. 2. Find the equation to the circle which touches the axis ofy
at a distance + 4 from the origin and cuts off an intercept 6 from the
axis of X.
Any circle is x^ + y'^ + 2gx + 2fy + c = 0.
This meets the axis of y in points given by
y^ + 2fy + c = 0.
The roots of this equation must be equal and each equal to 4, so
that it must be equivalent to {y - 4)^ = 0.
Hence 2/= -8, and c = 16.
The equation to the circle is then
x'^ + y^ + 2gx-8y + 16 = 0.
This meets the axis of x in points given by
x'^ + 2gx + l&:=0,
i.e. at points distant
-g+slT^^ and -g-Jg^lJG,
Hence Q = 2^g^-ie.
Therefore ^= ±5, and the required equation is
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x^ + y^±10x-8y + U = 0.
There are therefore two circles satisfying the given conditions.
This is geometrically obvious.
EXAMPLES. XVII.
Find the equation to the circle
1. Whose radius is 3 and whose centre is ( - 1, 2).
2. Whose radius is 10 and whose centre is ( - 5, - 6).
3. Whose radius is a + b and whose centre is {a, - h).
4. Whose radius is Ja^ - b'^ and whose centre is { - a, - 6).
Find the^ coordinates of the centres and the radii of the circles
whose equations are
5. x^ + y^-ix-8y = ^l. Q, Sx'^ + Sz/ - 5x- Qy + 4 = 0.
124 COORDINATE GEOMETRY. [ExS.
7. x^ + y^=k{x + k). 8. x^ + y'^ = 2gx-2fy.
9. Jl+m"^ {x^ + if) - 2cx - 2mcy = 0.
Draw tlie circles whose equations are
10. x^ + y^ = 2ay. H, 3x^ + 3y^=4x.
12. 5x'^ + 5y^=2x + 3y.
13. Find the equation to the circle which passes through the
points (1, - 2) and (4, - 3) and which has its centre on the straight
line 305 + 42/ = 7.
14. Find the equation to the circle passing through the points
(0, a) and (6, h), and having its centre on the axis of x.
Find the equations to the circles which pass through the points
15. (0, 0), (a, 0), and (0, 6). 16. (1, 2), (3, -4), and (5, -6).
17. (1,1), (2,-1), and (3, 2). 18. (5, 7), (8, 1), and (1, 3).
20. ABGB is a square whose side is <x; taking AB and AD as
axes, prove that the equation to the circle circumscribing the square is
21. Find the equation to the circle which passes through the
origin and cuts off intercepts equal to 3 and 4 from the axes.
22. Find the equation to the circle passing through the origin
and the points (a, 6) and (6, a). Find the lengths of the chords that
it cuts off from the axes.
23. Find the equation to the circle which goes through the origin
and cuts off intercepts equal to h and h from the positive parts of the
axes.
24. Find the equation to the circle, of radius a, which passes
through the two points on the axis of x which are at a distance & from
the origin.
Find the equation to the circle which
25. touches each axis at a distance 5 from the origin.
26. touches each axis and is of radius a,
27. touches both axes and passes through the point ( - 2, - 3).
28. touches the axis of x and passes through the two points
(1, -2) and (3, -4).
29. touches the axis of y at the origin and passes through the
point (6, c).
XVII.] TANGENT TO A CIRCLE. 125
30. touches the axis of a; at a distance 3 from the origin and
intercepts a distance 6 on the axis of y.
31. Points (1, 0) and (2, 0) are taken on the axis of x, the axes
being rectangular. On the line joining these points an equilateral
triangle is described, its vertex being in the positive quadrant. Find
the equations to the circles described on its sides as diameters.
32. If 1/ = mx be the equation of a chord of a circle whose radius is
a, the origin of coordinates being one extremity of the chord and the
axis of X being a diameter of the circle, prove that the equation of a
circle of which this chord is the diameter is
(1 + m2) (£c2 + 1/2) -2a{x + my) = 0.
33. Find the equation to the circle passing through the points
(12, 43), (18, 39), and (42, 3) and prove that it also passes through
the points ( - 54, - 69) and ( - 81, - 38).
34. Find the equation to the circle circumscribing the quadrilateral
formed by the straight lines
2x + 3y = 2, dx-2y = 4:, x + 2y = 3, and 2x-y = B.
35. Prove that the equation to the circle of which the points
{x^ , y^) and {x^ , y^) ^^^ the ends of a chord of a segment containing an
angle 6 is
± cot 9 [{x - Xj) {y - 2/2) - {x -x^) (y - y^)] = 0.
36. Find the equations to the circles in which the line joining the
points {a, h) and {b, - a) is a chord subtending an angle of 45° at any
point on its circumference.
148. Tangent. Euclid in his Book III. defines the
tangent at any point of a circle, and proves that it is always
perpendicular to the radius drawn from the centre to the
point of contact.
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From this property may be deduced the equation to the
tangent at any point (x\ y') of the circle x^ ^-y^ - o?.
For let the point P (Fig. Art. 139) be the point
The equation to any straight line passing through T is,
by Art. 62,
Also the equation to OP is
126 COORDINATE GEOMETRY.
The straight lines (1) and (2) are at right angles, i.e. the
line (1) is a tangent, if
mx^ = -l, (Art. 69)
y
Substituting this value of m in (1), the equation of the
tangent at {x\ y) is
t.e. xx+yy - x'^ + y (oj.
But, since {x, y) lies on the circle, we have x^ + y"^ - a^,
and the required equation is then
XX' + yy' = a2.
149. In the case of most curves it is impossible to
give a simple construction for the tangent as in the case of
the circle. It is therefore necessary, in general, to give a
different definition.
Tangent. Def. Let F and Q be any two points, near
to one another, on any curve.
Join TQ \ then FQ is called a
secant.
The position of the line PQ when
the point Q is taken indefinitely close
to, and ultimately coincident with, the
point F is called the tangent at F. X ^
The student may better appreciate ^^--,___^^
this definition, if he conceive the curve
to be made up of a succession of very small points (much
smaller than could be made by the finest conceivable drawing
pen) packed close to one another along the curve. The
tangent at F is then the straight line joining F and the
next of these small points.
150. To find the equation of the tangent at the imint
{x\ y) of the circle x^ + y^ = a^.
EQUATION TO THE TANGENT. 127
Let P be the given point and Q a point (x\ y") lying on
the curve and close to P.
The equation to FQ is then
Since both (cc', y) and {x\ y") lie on the circle, we have
By subtraction, we have
i. e. {x" - x') {x" + x') + (y" - y') (y" + y') ^ 0,
y - y X + x
X - X y + y
Substituting this value in (1), the equation to FQ is
Now let Q be taken very close to F, so that it ulti-
mately coincides with P, i, e. put x" = x and y" = y.
Then (2) becomes
2x
^y
The required equation is therefore
xx' + yy' = a2 (3).
It will be noted that the equation to the tangent
found in this article coincides with the equation found
from Euclid's definition in Art. 148.
Our definition of a tangent and Euclid's definition there-
fore give the same straight line in the case of a circle,
151. To obtain the equation of the tangent at any point
(x'y y') lying on the circle
x^ + y'^ + 2gx + 2fy + c = 0.
128 COORDINATE GEOMETRY.
Let P be the given point and Q a point (x'\ y") lying on
the curve close to P.
The equation to PQ is therefore
Since both (x\ y) and (cc", y") lie on the circle, we have
and x"^ + y"' + 2gx" + 2fy" + c^^ (3).
By subtraction, we have
^- _ x'^ + y"^ _ y- + 2g (x" - x') + 2/(y" - y') = 0,
i. e. (x" - x) {x" + x' + 2g) + (y" - y') {y" + y' + 2/) = 0,
y - y X + X •{• Zg
X - X y +y +2/
Substituting this value in (1), the equation to PQ be-
comes
Now let Q be taken very close to P, so that it ultimately
coincides with P, i. e. put x" - x and y" = y'.
The equation (4) then becomes
^. e. y {y +/) + 0^(03' + ^) = y' {y' +f)+x {x' + g)
to fO / /• /
= a:^ + 2/- + p'aj +fy
by (2).
This may be written
which is the required equation.
152. The equation to the tangent at (x', y') is there-
fore obtained from that of the circle itself by substituting
XX for x^y yy' for y^, x ■{■ x' for 2a;, and y -V y for 2y.
INTERSECTIONS OF A STRAIGHT LINE AND A CIRCLE. 129
This is a particular case of a general rule Avhich will be
found to enable us to write down at sight the equation to
the tangent at ix\ y') to any of the curves with which we
shall deal in this book.
153. Points of intersection, in general, of the straight
line
y^mx -\- G (1),
with the circle x' + y^ = d^ (2).
The coordinates of the points in which the straight line
(1) meets (2) satisfy both equations (1) and (2).
If therefore we solve them as simultaneous equations
we shall obtain the coordinates of the common point or
points.
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Substituting for y from (1) in (2), the abscissae of the
required points are given by the equation
The roots of this equation are, by Art. 1, real, coinci-
dent, or imaginary, according as
(2mc)^ - 4 (1 + m^) (c^ - a^) is positive, zero, or negative,
i.e. according as
a? {\ -\- m^) - c^ is positive, zero, or negative,
i.e. according as
In the figure the lines marked I, II, and III are all
parallel, i.e. their equations all have the same "r/i."
L. 9
130 COORDINATE GEOMETRY.
The straight line I corresponds to a value of c^ which
is <a^ (1 + oYv^) and it meets the circle in two real points.
The straight line III which corresponds to a value of c^,
> a^ (1 + m^), does not meet the circle at all, or rather, as in
Art. 108, this is better expressed by saying that it meets
the circle in imaginary points.
The straight line II corresponds to a value of c', which
is equal to a^ (1 + m^), and meets the curve in two coincident
points, i.e. is a tangent.
154. We can now obtain the length of the chord inter-
cepted by the circle on the straight line (1). For, if x^ and
X2 be the roots of the equation (3), we have
2mc _, c^ - cC-
Hence
^1 - ^2 = Jip^i + ^2)^ - ^x^x^ = -z "2 Jm^c^ - {(f - a^) (1 + m^)
J. "r 7Yh
1 +m^
If 2/1 and 2/2 ^^ t^6 ordinates of Q and R we have, since
these points are on (1),
2/1 - 2/2 = {mx-^ + <^) "~ (^^2 + c) = rri {x^ - x^.
Hence
QR = J(yi - y^Y + K - oc^y = Jl+ m' (x^^ - x.^
y
1 + 7n^
In a similar manner we can consider the points of inter-
section of the straight line y = mx + k with the circle
a;^ + 2/2 + 2yx + 2fy + c = 0.
155. The straight line
y = mx + ajl + m^
is always a. tangent to the circle
2 o
EQUATION TO ANY TANGENT. 131
As in Art. 153 the straight line
y = mx + G
meets the circle in two points which are coincident if
But if a straight line meets the circle in two points
which are indefinitely close to one another then, by Art.
149, it is a tangent to the circle.
The straight line y = mx + c is therefore a tangent to the
circle if
G^a J\ + 772^,
i.e. the equation to any tangent to the circle is
y = mx + a Vl + m2 (1).
Since the radical on the right hand may have the + or -
sign prefixed we see that corresponding to any value of in
there are two tangents. They are marked II and IV in
the figure of Art. 153.
156. The above result may also be deduced from the equation
(3) of Art. 150, which may be written
x' a2
y y
x'
Put - , = 'm, so that x'= -my' , and the relation .r'- + ?/'2 = a^ gives
The equation (1) then becomes
y = mx + afjl + m\
This is therefore the tangent at the point whose coordinates are
~ma . a
and
Vi + w2 7i +
157. If we assume that a tangent to a circle is always perpen-
dicular to the radius vector to the point of contact, the result of
Art. 155 may be obtained in another manner.
For a tangent is a line whose perpendicular distance from the
centre is equal to the radius.
9 - 2
132 COORDINATE GEOMETRY.
The straight line y=mx + c will therefore touch the circle if the
perpendicular on it from the origin be equal to a, i.e. if
i.e. if c = a /sjl + m^.
This method is not however applicable to any other curve besides the
circle.
158. Ex. Find the equations to the tangents to the circle
x'^ + y^-6x + Ay = 12
which are parallel to the straight line
4a;+3i/ + 5 = 0.
Any straight line parallel to the given one is '
The equation to the circle is
(a; -3)2 + (2/ + 2)2 =52.
The straight line (1), if it be a tangent, must be therefore such
that its distance from the point (3, - 2) is equal to ±5.
Hence ^l7^±f=:±5 (Art. 75),
V4H32
so that (7= -6=t25 = 19or -31.
The required tangents are therefore
4a; + 32/ + 19 = 0 and 4;r + 3?/-31 = 0.
159. Normal. Def. The normal at any point P of
a curve is the straight line which passes through P and is
perpendicular to the tangent at P.
To find the equation to the normal at the j^oint (x', y'^ of
(1) the circle
and (2) the circle
X- + y^ + Igx + 2/2/ + c = 0.
(1) The tangent at (.t', 3/') is
|
Internet Archive Historical Texts - Chunked (0001-1899)
TL;DR
- 163 million text chunks extracted from historical public-domain documents sourced from the Internet Archive
- Content dated 0001-1899, sorted by download popularity to prioritize high-quality, frequently accessed materials
- 2,445 Zstandard-compressed Parquet shards totaling ~217 GB on disk, ~594 billion characters uncompressed
- Optimized chunk size of ~3,600 characters (target: 4,000) for efficient language model training
- Cleaned OCR text with disclaimer removal, artifact filtering, and whitespace normalization
- Primarily English content with traces of other European languages
Quick Stats
| Metric | Value |
|---|---|
| Total text chunks | ~163,365,120 |
| Total characters | ~593,707,573,963 (593.7B) |
| Parquet shards | 2,445 |
| On-disk size (compressed) | 216.9 GB |
| Average chunk size | 3,634 chars |
| Median chunk size | 3,808 chars |
| Chunk size range | 102 - 7,928 chars |
| Target chunk size | 4,000 chars |
| Primary language | English (~97%) |
Dataset Description
Overview
This dataset contains chunked historical texts from the Internet Archive, preprocessed for language model training. The source materials span from year 0001 to 1899 and were selected based on download counts to ensure quality and relevance. Long documents have been split into manageable chunks of approximately 4,000 characters each, making the dataset ideal for:
- Pre-training language models on historical English text
- Fine-tuning models for historical document understanding
- Historical NLP research and analysis
- OCR quality assessment and improvement
Chunk Statistics
| Percentile | Chunk Size (chars) |
|---|---|
| P25 | 3,497 |
| P50 (Median) | 3,808 |
| P75 | 3,948 |
| P90 | 3,987 |
| P95 | 3,996 |
| P99 | 4,002 |
The distribution shows most chunks cluster around the 4,000 character target, with a small tail of shorter chunks from the end of documents or naturally brief sections.
Data Format
Each Parquet shard contains:
- Column:
text(string) - Row group size: 1,024 rows
- Compression: Zstandard (level 3)
- Typical chunks per shard: 60,000-70,000
Files are named sequentially: shard_00000.parquet through shard_02444.parquet
Preprocessing Pipeline
The dataset underwent extensive cleaning to maximize quality:
- Disclaimer Removal: Stripped Internet Archive, Google Books, and JSTOR boilerplate
- OCR Artifact Filtering: Removed page numbers, annotations, and noise patterns
- Text Normalization: Standardized whitespace, fixed common OCR errors
- Quality Filters:
- Minimum chunk length: 100 characters
- Chunks split at natural paragraph boundaries when possible
- Aggressive deduplication of boilerplate content
- Chunking Strategy:
- Target size: 4,000 characters
- Smart splitting on paragraph breaks to preserve context
- Large paragraphs split on sentence boundaries
- Minimum chunk size enforced to avoid fragmentary text
Content Examples
The following are real, unedited samples from the dataset showing the variety of content and text quality:
Example 1: Literary - Historical Poetry (Edmund Spenser's Epithalamion)
Ring ye the bells, ye yong men of the tow no.
And leave your wonted labors for this day :
This day is holy - do ye write it downe,
That ye for ever it remember may, -
This day the sun is in his chiefest hight,
With Barnaby the bright.
From whence declining daily by degrees,
He somewhat loseth of his heat and light,
When once the Crab behind his back he see.«
But for this time it ill -ordained was
POEMS OF LOVE.
To choose the longest day in all the yeare,
A.nd shortest night, when longest fitter
weare ;
Yet never day so long but late would passe.
Ring ye the hells, to make it weare away,
And bonfires make all day ;
And daunce about them, and about them sing,
That all the woods may answer, and theyr echo ring.
Example 2: Literary Criticism - Analysis of Historical Writing Style
Often does he declare that he purposely varies his diction,
lest the reader should be disgusted by its sameness; anx-
iously careful to avoid repetition, even in the structure of his
phrases. It may be said, however, that generally, in his
earlier works, (for he was apparently very young when he
wrote his History of the Kings,) his style is rather laboured ;
though, perhaps, even this may have originated in an anxiety
that his descriptions should be full ; or, to use his own ex-
pression, that posterity should be wholly and perfectly in-
formed. That his diction is liighly antithetical, and his
sentences artfully poised, will be readily allowed; and per-
haps the best index to his meaning, where he may be occa-
sionally obscure, is the nicely-adjusted balance of his phrase.
That he gradually improved his style, and in riper years,
where he describes the transactions of his own times, became
terse, elegant, and polished, no one will attempt to dispute.
Example 3: Religious/Historical - Biography of Father Gallitzin
There was one person, much nearer the scene of action,
who alone appears to have had the necessary force and firm-
ness, the indomitable courage, and the all-mastering will to
face and to thoroughly conquer the storm the others dared
not meet; it was the place for Father Gallitzin's immense
faith and magnificent spirit; he alone appears endowed with
that lion like nature, fortified by long trials and experienced
in the wickedness of rebellious man,-, inspired and strength-
ened beyond all human force by the battle cry forever in his
ears : God wills it, which fears not, single handed, to meet a
legion of enemies. But a superior wisdom so ordered it that
the evil thing should have its day and run its course.
Example 4: Historical Legal Text - Irish and European Land Tenure
The Irish, tenures throw considerable light upon many ob-
onThoseof* scure points in the tenures of the rest of Europe in medieval
Europe, times ; for instance there can be no doubt that hereditary tenan-
cies, the " Erbpacht" of the Germans, and the Emphyteusis of
the later Roman Empire, co-existed all through the middle ages
in Italy, Germany, France, and Flanders, with a system of
villenage analogous to the Irish Fuidirship. Marini and
Mabillon mention tenants of the former class under the name
of libellarii from the sixth to the thirteenth centuries;351
they were numerous also in Germany in the ninth century.
The greater part of the occupiers of land in France in the ninth
century were in the position of Ceiles, holding by limited ser-
vice, is proved by documents forbidding the raising of rents.
The serfs proper we know held their land in the greater part
of Germany as an inheritance from the thirteenth century.355
Example 5: Biographical - Judge Sir John Williams
{)ublic at large became aware of his match-
ess talents in that branch of an advocate's
duty. Professional success followed "the
Queen's trial." Mr. Williams then got into
Parliament, sitting for Lincoln, Winchel-
sea, and Ilchester, on the Liberal interest,
and distinguished his Parliamentary career
by his advocacy of Chancery Reform. A
change of the Ministry at length procured
for him that professional position to which
he had been for some years fairly entitled.
He received a silk gown, and soon after
the accession of William IV her Majesty,
now Queen Dowager, appointed him her
Attorney-General. In Feb. 1834 he be-
came one of the Barons of the Exche-
quer, and having sat in that court only
one term was transferred to the Court of
King's Bench, where he remained until
the period of his lamented death.
Example 6: Technical - Railroad Coupler Specifications (1899)
3. To propose specifications for couplers. This part of the subject has received
very careful consideration. It has been difficult to reconcile the diametrically
opposite opinions which have been expressed by various railroad men and manu-
facturers. It is believed, however, that rigid specifications and tests will do much
to weed out the poorer makes of couplers at present being furnished, and it is rec-
ommended that in the future all couplers be purchased subject to the provisions of
the following standard specifications and tests.
A. B. &• C. R. R. CO.
Specifications for M. C. B. Automatic Car Couplers.
After September 1, 1899, aH M. C. B. automatic car couplers purchased by or
used in the construction of cars for the above-named company must meet the require-
ments of the following specifications.
Example 7: Theological - Church Doctrine Discussion
teacheth, that in this mystery there is not in the bread a substantial
but a sacramental change, according to the which the outward ele-
ments take the name of what they represent, and are changed in
such a sort that they still retain their former natural substance.
"The bread," saith he, "is made worthy to be honoured with the
name of the Flesh of Christ by the consecration of the priest, yet the
Flesh retains the proprieties of its incorruptible nature, as the bread
doth its natural substance. Before the bread be sanctified, we call
it bread; but, when it is consecrated by the divine grace, it deserves
to be called the Lord's Body, though the substance of the bread still
remains'."' When Bellarmine could not answer this testimony of
that great doctor, he thought it enough to deny that this epistle is
S. Chrysostom's'*: but both he and Possevin do vainly contend that
it is not extant among the works of Chrysostom.
Example 8: Literary - 19th Century Novel (Character Description)
Had she been less evenly balanced, had her
soul been less true, her heart less tender, she
might in time have frozen the woman complete!),
and crystallized into the artiste only - or - but
to think of Judith Moore sullying her wings is
sacrilege.
She was full of womanly tenderness and
womanly vanities. She had a thousand little
tricks of coquetry and as many balms to care
their smart. She took a good deal of satis fac
tion out of her pretty gowns and her finger
nails, and the contemplation of her little feet
becomingly shod had been known to dry her
tears. She was essentially the woman of the
past, the woman who created a " type " distinct
from man; the womanly woman, not the hybrid
creature of modern cultivation ; the woman of
romance.
Example 9: Historical Records - Marriage Registry (OCR artifacts visible)
Michael Darey and Aan Cusack.
George Omensetter and Margaret Sainer.
Wilhehn Denzel, wid% and Elizabeth Jansou.
Thomas Butbis and Pattj" Post.
Peter Lengfelder and Barbara Birkenbeiler.
Andrew McFarlene, wid% and Sarah Lakorn, wid.
Carl Himmelreich and Susanna Funck.
John Tallentire and Elizabeth Shade.
Joseph Bolton and Sarah Hofty.
George Fried. Wendt and Sarah Charlotte Eichbaum.
Jacob Chur, wid^ and Wendeling Margar. Dorneck.
Adam Hyner and Elizabeth Wears (Wehrs).
Gideon Cox and Susanna Shevely.
Note: "wid%" artifacts are OCR errors for "widower" or "widow"
Example 10: Reference/Directory - Business Listings (1899)
STORES, OFFICES AND LOFTS.
BUFFALO, N Y. - Wood & Bradne/
Mutual Life Bldg. Buffalo, have plans In
progress for remodeling the 5-sty brick
business block, 35x200, to include stores,
offices, arcade billiard and pool room, at
319 Main st, through from Main to Wash-
ington sts. for Dr. and Conrad E. Witt-
laufer, 1234 Delaware av, Buffalo, owne'\
ROCHESTER, N. Y. - Gordon & Madden.
300 Sibley Block, Rochester, have work-
ing plans in progress for addition to the
2-sty brick and tile School No. 27, in Cen-
tral Park, cor 1st st, for the City of Roch-
ester, Board of Education, J. S. Mullen, 37
Exchange st, Rochester, owner. Cost, $16,000.
Example 11: Historical Encyclopedia - Roman Emperor Diocletian
DiocLrnixcs. Caios Valerius Jovtus, a cele-
brated Roman emperor, born of an obscure family ia
Dalmatia, at the town of Dioclea or Doclea, from
which town be derived bis first name, which was
probably Doclea, afterward lengthened to the more
harmonious Greek form of Dioclea, and at length,
after his accession to the empire, to the Roman form
of Diocleti&nus. He likewise, on this occasion, as-
sumed the patrician name of Valerius. Some, how-
ever, make him to nave been born at Salona. Hie
birth year also i» differently given. The common
account says 245 A.D., but other statements make
bin tea years older. He waa first a common soldier,
and by merit and success gradually rose to rank.
Example 12: Poetry - Personification of Winter
Some call me their foe, but I hone and intend
To make it appear, I am truly your friend ;
You may think mv deportment is furly and bluff,
But I mean it for good when I handle you rough.
My fnow when descending it covers your fields,
The beft of manuring confcftdly yields.
While its fmooth fhiuing furface aiiords you a fpace
For your fleighs and your fledges to drive at full chace.
My ice, how reviving in heat does it feem,
It cools all your liquors and fwteten.s your cream ;
On ^Etna's tall fummit 'lis gadier'd, and thence
O'er Italy does its refrefhment difpeuf*.
Note: This example shows period spelling conventions and some OCR artifacts (fnow=snow, fpace=space)
Usage
Loading the Dataset
Using PyArrow (Recommended)
import pyarrow.dataset as ds
import pyarrow.compute as pc
# Load all shards as a single dataset
dataset = ds.dataset("shard_*.parquet", format="parquet")
# Efficient streaming with multi-threading
scanner = dataset.scanner(
columns=["text"],
use_threads=True,
batch_size=4096
)
for batch in scanner.to_batches():
texts = batch["text"].to_pylist()
# Process texts...
Using Pandas
import pandas as pd
# Load a single shard
df = pd.read_parquet("shard_00000.parquet")
print(df.head())
# Load all shards (requires sufficient RAM)
import glob
files = sorted(glob.glob("shard_*.parquet"))
df = pd.concat([pd.read_parquet(f) for f in files])
Using HuggingFace Datasets
from datasets import load_dataset
# Load from local path
dataset = load_dataset("parquet", data_files="shard_*.parquet")
# Iterate efficiently
for example in dataset["train"]:
text = example["text"]
# Process text...
Computing Statistics
Quick script to verify the dataset:
import pyarrow.dataset as ds
import pyarrow.compute as pc
dataset = ds.dataset("shard_*.parquet", format="parquet")
scanner = dataset.scanner(columns=["text"], use_threads=True)
count = 0
total_chars = 0
for batch in scanner.to_batches():
lengths = pc.utf8_length(batch["text"])
count += batch.num_rows
total_chars += pc.sum(lengths).as_py()
print(f"Chunks: {count:,}")
print(f"Characters: {total_chars:,}")
print(f"Avg size: {total_chars // count:,} chars")
Language Distribution
Based on sampling 200 documents across the dataset using langdetect:
| Language | ISO Code | Percentage |
|---|---|---|
| English | en | ~97% |
| French | fr | ~1% |
| Dutch | nl | ~0.5% |
| Other European | various | ~1.5% |
Known Limitations
- OCR Errors: Despite cleaning, some OCR artifacts remain, especially in older or lower-quality scans
- Historical Spelling: Texts preserve original spelling and grammar, which may differ from modern conventions
- Content Bias: Download-based sorting skews toward popular topics (legal texts, classics, frequently referenced works)
- No Metadata: Author, title, publication year, and other bibliographic data are not included in the chunks
- Language Imbalance: Heavily English-dominant due to Internet Archive's collection composition
- Chunk Boundaries: While optimized for readability, some chunks may split mid-thought
Ethical Considerations
- All source materials are from Internet Archive's public domain collection
- Users should verify the public domain status in their jurisdiction before commercial use
- Historical texts may contain outdated or offensive viewpoints that do not reflect modern values
- Recommended for research and model training; review outputs before production deployment
Citation
If you use this dataset, please cite:
@dataset{internet_archive_chunked_1899,
title={Internet Archive Historical Texts - Chunked (0001-1899)},
year={2025},
publisher={Internet Archive},
note={Processed and chunked from Internet Archive public domain texts dated 0001-1899, sorted by download count}
}
Please also acknowledge the Internet Archive for the source materials:
@misc{internetarchive,
title={Internet Archive},
howpublished={\url{https://archive.org}},
note={Accessed: 2025}
}
Technical Details
Shard Creation
- Row group size: 1,024 chunks per group
- Compression: Zstandard level 3 (balance of speed and compression)
- Target shard size: ~250M characters per shard
- Write batch size: 10,000 rows
Performance Tips
- Enable Arrow memory mapping for zero-copy reads on supported filesystems
- Use
use_threads=Truein PyArrow scanners to leverage multi-core CPUs - Stream batches with
to_reader()instead of materializing entire dataset - For sampling, use
pc.utf8_slice_codeunits()to avoid loading full multi-megabyte chunks
Storage Recommendations
- SSD/NVMe: Recommended for training pipelines
- Compression ratio: ~2.7x (594GB uncompressed → 217GB compressed)
- Random access: Each shard is independently readable
Version History
- v1.0 (2025): Initial release with 2,445 shards, 163M chunks, 594B characters
License
This dataset contains materials from the Internet Archive's public domain collection. While the source materials are in the public domain, users should:
- Verify the legal status of specific works in their jurisdiction
- Comply with Internet Archive's Terms of Use
- Review individual works for any usage restrictions
The preprocessing scripts and this dataset compilation are provided as-is for research and educational purposes.
Suitability for Historical Language Modeling
Would this dataset create an LLM representative of a person from 1899?
Short answer: Partially, with significant caveats.
Strengths:
- ✅ Authentic historical vocabulary and phrasing - The texts use period-appropriate language, spelling conventions, and sentence structures
- ✅ Diverse subject matter - Includes legal documents, literature, religious texts, biographical materials, scientific works, poetry, and commercial records
- ✅ Representative of educated/literate discourse - Reflects how educated people wrote in the 19th century and earlier
- ✅ Rich contextual knowledge - Contains historical events, social norms, and cultural references from the period
Limitations:
- ❌ Not conversational/spoken language - These are published works, not everyday speech or personal correspondence
- ❌ Heavy OCR artifacts - Despite cleaning, remnants like "wid%" (widow), formatting errors, and garbled text persist
- ❌ Bias toward formal/academic writing - Overrepresents legal texts, religious works, and scholarly materials
- ❌ Limited personal voice - Lacks diaries, letters, informal notes that would reflect casual 1899 communication
- ❌ Skewed by download popularity - Popular classics and reference works are overrepresented
- ❌ Temporal mixing - Contains texts from across 1,900 years (0001-1899), not just the 1890s
Text Quality Assessment:
From sampled chunks, the dataset contains:
- Legal/Administrative (30-40%) - Court cases, legislation, property records, business directories
- Literary (20-30%) - Poetry (Shakespeare, Spenser), novels, historical narratives
- Religious/Theological (15-20%) - Sermons, biblical commentary, church records
- Historical/Biographical (10-15%) - Historical accounts, biographical sketches, chronicles
- Technical/Reference (10-15%) - Specifications, mathematical tables, encyclopedic entries
- Marriage/Death Records (5-10%) - Lists of names and vital statistics
OCR Quality Issues Observed:
- Formatting artifacts: "wid%" for widow, "wull" for will, "tke" for the
- Table data corruption (numbers in columns)
- Character substitution: "honour" (correct historical spelling vs. OCR error - hard to distinguish)
- Spacing issues and paragraph breaks
Recommendation for Creating a "1899 Person" LLM:
This dataset would be most effective when:
Combined with other sources:
- Personal letters and diaries from 1890s
- Newspaper articles (more conversational)
- Period fiction dialogue
- Oral history transcriptions
Used with heavy post-processing:
- Additional OCR error correction
- Filtering to focus on 1880s-1899 materials only
- Weighting toward more conversational genres
- Removing heavily corrupted chunks
Contextualized as formal written English:
- Best for modeling formal 19th-century writing style
- Good for historical knowledge and cultural references
- Not ideal for casual conversation simulation
Final Assessment: This dataset provides excellent historical language patterns and knowledge but would produce an LLM that writes like a 19th-century author or scholar, not necessarily how an average person from 1899 would speak. For that, you'd need substantial supplementary data from informal sources.
The dataset's value lies in teaching historical vocabulary, grammatical structures, cultural context, and the formal register of 19th-century English - making it a strong foundation that requires augmentation for truly representative historical persona modeling.
Contact & Contributions
For issues, suggestions, or questions about this dataset:
- Check the Internet Archive for source material questions
- Report data quality issues through the dataset repository
Acknowledgments
- Internet Archive for preserving and providing access to historical texts
- PyArrow team for high-performance data processing tools
- The open-source community for OCR and text processing tools
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