Copyright
S. L. (Sidney Luxton) Loney.

The elements of coordinate geometry online

. (page 6 of 26)
Online LibraryS. L. (Sidney Luxton) LoneyThe elements of coordinate geometry → online text (page 6 of 26)
Font size
QR-code for this ebook


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

\-a,a {— + — + — + U .

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.



81




Let AB be the fixed base of the triangle j take its
middle point 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 a 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 and -; = tan ^.



h + a h — a

Substituting these values in (1), we have

1{h + a)k



k



h + a



h



a



1



(h+ay-k^'



\h + a/

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 ; 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

Ah + Bk + C = (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.



83



Let the triangle be OAB and take 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



- + — = 1
a Xb






(1).



(2)-



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



%.e.



a b'

This then is the equation to the locus of P.
always lies on the straight line

b
y = — x.



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 ^•

A. + 1 \+ 1

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 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

-= -^ and - = y^, .

r C r C



THE STRAIGHT LINE. PROBLEMS. 85

1 1

If the angle XOE be 6 the values of ^r^ and ;=r-5 are

therefore

^ cos ^ + ^ sin A' cos $ + B' sin 6
-^ and ~ .

We therefore have

2 _ ^cos^ + .gsin^ A' cos + B' sin 6
0P~~ C C'

fA A'\ a (^ ^'\ ' a

The equation to the locus of P is therefore, on again
transforming to Cartesian coordinates,

_ (A A'\ (B B\

and this is a fixed straight line.



EXAMPLES. XL

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 ; 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 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 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 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 ; 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 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 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.

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 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 = (1).

This equation is equivalent to

{y-x){y-2x) = (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

2/-^ = (3),

and also by the points which satisfy

y-2x = (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 = 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'^ ^ (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 = (2),

and ax + hy — y J li^ — ab = (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 = 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.

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- = (1).

Let the separate equations to the two lines be

y—miX = and y~ni^x — (2),

so that (1) must be equivalent to

b (y — mjx) (y — tUox) = (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 be the angle between the straight lines (2) we
have, by Art. 66,

tan 6 = ^^ ~ ^^^^ =_- n/(^i + ^2)^ - ^rn-jn ^
1 + m^m^ 1 + m-{nio,



J



F~T 2Vh2-ab



.(4).



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 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 = and Qx^ + llxy — 6^- —

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 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.



92



COORDINATE GEOMETRY.



112. Tojind the equation to the straight lines bisecting
the angle between the straight lines given by

ax'' + 2hxy + by^ = (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

tan 2^ = tan {0^ + ^o) = r-^-a^ — ^ = " I '

' 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^

a; 2xy



*? *> 9. 5

a?"
x2 - y2 xy

I.e. Z - = -r— .

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
-2fe , „ .,. „ /a
6



{x^-y-') + 2{~-ljxy = 0,



x^-y"^ _ xy
I.e. — — ^-— .

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.

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 + cos2 ^j _ 2xy tan d + y^ sin^ ^ =

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^] =
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 == (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 = (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=..
/^/h^ — ab

The relation (7) then gives

Aac = 4:AB = {A+B)^-{A-By-

~^^ ^ h^-ab '
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

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 =
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-
= - 1875 + ^^-^^ = 0.
The equation therefore represents two straight lines.
Solving it for x, we have

o,, 7y + 13 f7y + lBY_ 10y^-4.mj + S5 f ly + is y

^ +"^-12" + V ~ 2^-; - — 12 — + \2r~)

_/2% - 43\2

~l 24 J •
. , , 7y + 13 _ 2 3y-43
•• ^+-2^-=^ - 24""'

2?/ - 7 - 5w + 5

I. e. x= -— — or — -. .

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 =
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,
i.e. (2a; + 3?/ + 4) (3a; + 4?/ + 5) = 0.

Taking the second value, the equation is

20a;2 + 57a;?/ + 40?/2 + ^fa a; + H-2/ + -t- = 0,
i.e. (4a; + o?/ + -V)(5a; + 8?/ + 10) = 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 =



Online LibraryS. L. (Sidney Luxton) LoneyThe elements of coordinate geometry → online text (page 6 of 26)