Exercise 6.4 - Chapter 6 Two Dimensional Analytical Geometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

$
\text { Ex } 6.4
$
Question 1.
Find the combined equation of the straight lines whose separate equations are $x-2 y-3=0$ and $x+y+$ $5=0$.
Solution:
Separate equations are $x-2 y-3=0 ; x+y+5=0$
So the combined equation is $(x-2 y-3)(x+y+5)=0$
$
\begin{aligned}
& x^2+x y+5 x-2 y^2-2 x y-10 y-3 x-3 y-15=0 \\
& \text { (i.e) } x^2-2 y^2-x y+2 x-13 y-15=0
\end{aligned}
$
Question 2.
Show that $4 x^2+4 x y+y^2-6 x-3 y-4=0$ represents a pair of parallel lines.
Solution:
Comparing this equation with $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$
we get $\mathrm{a}=4, \mathrm{~h}=\frac{4}{2}=2, \mathrm{~b}=1, \mathrm{~g}=-3, \mathrm{f}=-3 / 2, \mathrm{c}=-4$
The condition for the lines to be parallel is $\mathrm{h}^2-\mathrm{ab}=0$
Now $\mathrm{h}^2-\mathrm{ab}=2^2-(4)(1)=4-4=0$
$\mathrm{h}^2-\mathrm{ab}=0 \Rightarrow$ The given equation represents a pair of parallel lines.
Question 3.
Show that $2 x^2+3 x y-2 y^2+3 x+y+1=0$ represents a pair of perpendicular lines.
Solution:
Comparing the given equation with the general form $a=2, h=3 / 2, b=-2, g=3 / 2, f=1 / 2$ and $c=1$
Condition for two lines to be perpendicular is $a+b=0$. Here $a+b=2-2=0$
$\Rightarrow$ The given equation represents a pair of perpendicular lines.
Question 4.
Show that the equation $2 x^2-x y-3 y^2-6 x+19 y-20=0$ represents a pair of intersecting lines. Show further that the angle between them is $\tan ^{-1}(5)$.
Solution:

Comparing the given equation with general form we get $a=2, b=-3, c=-20, f=\frac{19}{2}$, $g=-3, h=-1 / 2$
The condition for the given equation to represent a pair of straight lines is
$
\begin{aligned}
& a b c+2 f g h-a f^2-b g^2-c h^2=0 \\
& \begin{aligned}
\text { Now LHS } & =a b c+2 f g h-a f^2-b g^2-c h^2 \\
& =(2)(-3)(-20)+2\left(\frac{19}{2}\right)(-3)\left(-\frac{1}{2}\right)-2\left(\frac{19}{2}\right)^2+3(-3)^2+20(-1 / 2)^2 \\
& =120+\frac{57}{2}-\frac{361}{2}+27+5=180 \frac{1}{2}-180 \frac{1}{2}=0
\end{aligned}
\end{aligned}
$
The given equation represents a pair of straight lines.
The angle between the pair of straight lines is given by $\tan \theta= \pm \frac{2 \sqrt{h^2-a b}}{a+b}$
(i.e) $\tan \theta= \pm \frac{2 \sqrt{\frac{1}{4}+6}}{2+(-3)}= \pm \frac{2 \sqrt{\frac{25}{4}}}{-1}=2 \times \frac{5}{2}=5 \Rightarrow \theta=\tan ^{-1}(5)$
Question 5.
Prove that the equation to the straight lines through the origin, each of which makes an angle $\alpha$ with the straight line $y=x$ is $x^2-2 x y \sec 2 \alpha+y^2=0$
Solution:
Slope of $y=x$ is $m=\tan \theta=1$
$\Rightarrow \theta=45^{\circ}$
The new lines slopes will be
$\mathrm{m}=\tan (45+\alpha)$ and $\mathrm{m}=\tan (45-\alpha)$
$\therefore$ The equations of the lines passing through the origin is given by
$y=\tan (45+\alpha) x$ and $y=\tan (45-\alpha) x$
(i.e) $y=\tan (45+\alpha) x=0$ and $y=\tan (45-\alpha) x=0$
The combined equation is $[y-\tan (45+\alpha) x][y-\tan (45-\alpha) x]=0$
$y^2+\tan (45+\alpha) \tan (45-\alpha) x^2-x y[\tan (45-\alpha)+\tan (45+\alpha)]=0$

(i.e) $y^2+\frac{1+\tan \alpha}{1-\tan \alpha} \times \frac{1-\tan \alpha}{1+\tan \alpha} x^2-x y\left[\frac{\sin (45-\alpha)}{\cos (45-\alpha)}+\frac{\sin (45+\alpha)}{\cos (45+\alpha)}\right]=0$
(i.e) $x^2+y^2-x y \frac{[\sin (45-\alpha) \cos (45+\alpha)+\cos (45-\alpha) \sin (45+\alpha)]}{\cos (45-\alpha) \cos (45+\alpha)}=0$
(i.e) $x^2+y^2-x y \frac{[\sin (45+45)]}{\frac{1}{2}[2 \cos (45-\alpha) \cos (45+\alpha)]}=0$ $x^2+y^2-\frac{2 x y}{\cos 90+\cos 2 \alpha}=0 \Rightarrow x^2-\frac{2 x y}{\cos 2 \alpha}+y^2=0$
(i.e) $x^2-2 x y \sec 2 \alpha+y^2=0$

Aliter:
Let the slopes of the lines be $m_1$ and $m_2$ where
$
\begin{aligned}
& m_1=\tan (45-\alpha)=\frac{1-\tan \alpha}{1+\tan \alpha} \text { and } m_2=\tan (45+\alpha)=\frac{1+\tan \alpha}{1-\tan \alpha} \\
& m_1+m_2=\frac{1-\tan \alpha}{1+\tan \alpha}+\frac{1+\tan \alpha}{1-\tan \alpha} \\
& =2\left(\frac{1}{\cos 2 \alpha}\right)=2 \sec 2 \alpha \\
& m_1 m_2=\frac{1-\tan \alpha}{1+\tan \alpha} \times \frac{1-\tan \alpha}{1-\tan \alpha}=1
\end{aligned}
$

Let the equation of lines passes through the origin
So the equations are $y=m_1 x=0$ and $y=m_2 x=0$
So the combined equations is $\left(y-m_1 x\right)\left(y-m_2 x\right)=0$
(i.e) $y^2-x y\left(m_1+m_2\right)+m_1 m_2 x=0$
(i.e) $y^2-x y(2 \sec \alpha)+x^2(1)=0$
(i.e) $y^2-2 x y \sec 2 \alpha+x^2=0$
Question 6.
Find the equation of the pair of straight lines passing through the point $(1,3)$ and perpendicular to the lines $2 x-3 y+1=0$ and $5 x+y-3=0$
Solution:
Equation of a line perpendicular to $2 x-3 y+1=0$ is of the form $3 x+2 y+k=0$.
It passes through $(1,3) \Rightarrow 3+6+\mathrm{k}=0 \Rightarrow \mathrm{k}=-9$
So the line is $3 x+2 y-9=0$
The equation of a line perpendicular to $5 x+y-3=0$ will be of the form $x-5 y+k=0$.
It passes through $(1,3) \Rightarrow 1-15+\mathrm{k}=0 \Rightarrow \mathrm{k}=14$
So the line is $\mathrm{x}-5 \mathrm{y}+14=0$.
The equation of the lines is $3 x+2 y-9=0$ and $x-5 y+14=0$
Their combined equation is $(3 x+2 y-9)(x-5 y+14)=0$
(i.e) $3 x^2-15 x y+42 x+2 x y-10 y^2+28 y-9 x+45 y-126=0$
(i.e) $3 x^2-13 x y-10 y^2+33 x+73 y-126=0$

Question 7.
Find the separate equation of the following pair of straight lines
(i) $3 x^2+2 x y-y^2=0$
(ii) $6(x-1)^2+5(x-1)(y-2)-4(y-2)^2=0$
(iii) $2 x^2-x y-3 y^2-6 x+19 y-20=0$
Solution:
(i) Factorising $3 x^2+2 x y-y^2$ we get
$
\begin{aligned}
& 3 x^2+3 x y-x y-y^2=3 x(x+y)-y(x+y) \\
& =(3 x-y)(x+y)
\end{aligned}
$
So $3 x^2+2 x y-y^2=0 \Rightarrow(3 x-y)(x+y)=0$
$
\Rightarrow 3 x-y=0 \text { and } x+y=0
$
(ii) $6(x-1)^2+5(x-1)(y-2)-4(y-2)^2=0$
$\Rightarrow 6\left(x^2-2 x+1\right)+5(x y-2 x-y+2)-4\left(y^2-4 y+4\right)=0$
(i.e) $6 x^2-12 x+6+5 x y-10 x-5 y+10-4 y^2+16 y-16=0$
(i.e) $6 x^2+5 x y-4 y^2-22 x+11 y=0$
Factorising $6 x^2+5 x y-4 y^2$ we get
$
\begin{aligned}
& 6 x^2-3 x y+8 x y-4 y^2=3 x(2 x-y)+4 y(2 x-y) \\
& =(3 x+4 y)(2 x-y)
\end{aligned}
$
So, $6 x^2+5 x y-4 y^2-22 x+11 y=(3 x+4 y+1)(2 x-y+m)$
Equating coefficient of $x \Rightarrow 3 \mathrm{~m}+21=-22$
Equating coefficient of $y \Rightarrow 4 \mathrm{~m}-1=11$
Solving (1) and (2) we get $1=-11, \mathrm{~m}=0$
So the separate equations are $3 x+4 y-11=0$ and $2 x-y=0$
(iii) $2 x^2-x y-3 y^2-6 x+19 y-20=0$
Factorising $2 x^2-x y-3 y^2$ we get
$2 x^2-x y-3 y^2=2 x^2+2 x y-3 x y-3 y^2$
$=2 \mathrm{x}(\mathrm{x}+\mathrm{y})-3 \mathrm{y}(\mathrm{x}+\mathrm{y})=(2 \mathrm{x}-3 \mathrm{y})(\mathrm{x}+\mathrm{y})$
$\therefore 2 x^2-\mathrm{xy}-3 \mathrm{y}^2-6 \mathrm{x}+19 \mathrm{y}-20=(2 \mathrm{x}-3 \mathrm{y}+1)(\mathrm{x}+\mathrm{y}+\mathrm{m})$
Equating coefficient of $x 2 m+1=-6$
Equating coefficient of $y-3 m+1=19$
Constant term $-20=1 \mathrm{~m}$
Solving (1) and (2) we get $1=4$ and $\mathrm{m}=-5$ where $1 \mathrm{~m}=-20$.
So the separate equations are $2 x-3 y+4=0$ and $x+y-5=0$

Question 8.
The slope of one of the straight lines $a x^2+2 h x y+b y^2=0$ is twice that of the other, show that $8 h^2=9 a b$.

Solution:
$
a x^2+2 h x y+b y^2=0
$
We are given that one slope is twice that of the other.
So let the slopes be $\mathrm{m}$ and $2 \mathrm{~m}$.
Now sum of the slopes $=m+2 m$
$
=3 m=-\frac{2 h}{b} \Rightarrow m=-\frac{2 h}{3 b}
$
Product of the slopes $=m \times 2 m=2 m^2=a / b$
$
\Rightarrow m^2=\frac{a}{2 b}
$
Eliminating $m$ from (A) and (B) we get $\left(\frac{-2 h}{3 b}\right)^2=\frac{a}{2 b}$
(i.e) $\frac{4 h^2}{9 b^2}=\frac{a}{2 b}$
$
8 h^2=9 a b
$

Question 9.
The slope of one of the straight lines $a x^2+2 h x y+b y^2=0$ is three times the other, show that $3 h^2=4 a b$.

Solution:
Let the slopes be $\mathrm{m}$ and $3 \mathrm{~m}$.
$
\begin{aligned}
& \text { Now } m+3 m=4 m=-\frac{2 h}{b} \\
& \Rightarrow m=-\frac{2 h}{4 b}=-\frac{h}{2 b} \\
& m \times 3 m=\frac{a}{b} \Rightarrow 3 m^2=\frac{a}{b} \Rightarrow m^2=\frac{a}{3 b}
\end{aligned}
$
Eliminating $m$ from (1) and (2)
we get $\quad\left(-\frac{h}{2 b}\right)^2=\frac{a}{3 b} \Rightarrow \frac{h^2}{4 b^2}=\frac{a}{3 b} \Rightarrow 3 h^2=4 a b$
Question 10 .
A $\triangle O P Q$ is formed by the pair of straight lines $x^2-4 x y+y^2=0$ and the line $P Q$. The equation of $P Q$ is $x+y-2=0$. Find the equation of the median of the triangle $\triangle O P Q$ drawn from the origin $O$.
Solution:
Equation of pair of straight lines is $x^2-4 x y+y^2=0$
Equation of the given line is $\mathrm{x}+\mathrm{y}-2=0 \Rightarrow \mathrm{y}=2-\mathrm{x}$
$(2)$
On solving (1) and (2) we get $x^2-4 x(2-x)+(2-x)^2=0$
(i.e) $x^2-8 x+4 x^2+4+x^2-4 x=0$
(i.e) $6 x^2-12 x+4=0$
$(\div$ by 2$) 3 x^2-6 x+2=0$

$
\begin{aligned}
& x=\frac{6 \pm \sqrt{36-24}}{6}=\frac{6 \pm \sqrt{12}}{6} \\
& x=\frac{6 \pm 2 \sqrt{3}}{6}==\frac{3 \pm \sqrt{3}}{3}=1 \pm \frac{1}{\sqrt{3}} \\
& x=1+\frac{1}{\sqrt{3}}, 1-\frac{1}{\sqrt{3}}
\end{aligned}
$
When $x=1+\frac{1}{\sqrt{3}}, y=2-1-\frac{1}{\sqrt{3}}=1-1 / \sqrt{3}$
When $x=1-1 / \sqrt{3}, y=2-1+\frac{1}{\sqrt{3}}=1+\frac{1}{\sqrt{3}}$
$
\therefore \mathrm{P}=\left(1+\frac{1}{\sqrt{3}}, 1-\frac{1}{\sqrt{3}}\right) ; \mathrm{Q}=\left(1-\frac{1}{\sqrt{3}}, 1+\frac{1}{\sqrt{3}}\right)
$
Mid point of $\mathrm{PQ}$ is
$
\left(\frac{1+\frac{1}{\sqrt{3}}+1-\frac{1}{\sqrt{3}}}{2}, \frac{1-\frac{1}{\sqrt{3}}+1+\frac{1}{\sqrt{3}}}{2}\right)=(1,1)
$

Now the equation of the median through $(0,0)$ and $(1,1)$ is $\frac{y-1}{0-1}=\frac{x-1}{0-1}$
(i.e) $\frac{y-1}{-1}=\frac{x-1}{-1} \Rightarrow y-1=x-1$
(i.e) $x-y=0 \Rightarrow y=x$
Question 11.
Find $p$ and $q$, if the following equation represents a pair of perpendicular lines $6 x^2+5 x y-p y^2+7 x+q y$ $-50$
Solution:
$6 x^2+5 x y-p y^2+7 x+q y-50$
The given equation represents a pair of perpendicular lines
$\Rightarrow$ coefficient of $\mathrm{x}^2+$ coefficient of $\mathrm{y}^2=0$
(i.e) $6-p=0 \Rightarrow p=6$
Now comparing the given equation with the general form $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$
we get $a=6, b=-6$ and $c=-5, f=q / 2, g=7 / 2$ and $h=5 / 2$

The condition for the general form to represent a pair of straight lines is $a b c+2 f g h-a f^2-\mathrm{bg}^2-\mathrm{ch}^2=0$ $($ i.e $)(6)(-6)(-5)+2\left(\frac{q}{2}\right)\left(\frac{7}{2}\right)\left(\frac{5}{2}\right)-6\left(\frac{q^2}{4}\right)+6\left(\frac{49}{4}\right)+5\left(\frac{25}{4}\right)=0$
$
\text { (i.e) } \begin{aligned}
& 180+\frac{35 q}{4}-\frac{6 q^2}{4}+\frac{294}{4}+\frac{125}{4}=0 \\
& \frac{-6 q^2}{4}+\frac{35 q}{4}+\frac{720+294+125}{4}=0 \\
&-6 q^2+35 q+1139=0
\end{aligned}
$
Changing the sign throughout we get $6 q^2-35 q-1139=0$
$
\begin{aligned}
& q=\frac{35 \pm \sqrt{1225+4(6)(1139)}}{2(6)} \\
& q=\frac{35 \pm \sqrt{28561}}{12} \\
& q=\frac{35 \pm 169}{12} \\
& q=\frac{35+169}{12}, \frac{35-169}{12} \\
& q=17 \text { or }-\frac{67}{6} \text { and } p=6
\end{aligned}
$
Question 12.
Find the value of $\mathrm{k}$, if the following equation represents a pair of straight lines. Further, find whether these lines are parallel or intersecting, $12 x^2+7 x y-12 y^2-x+7 y+k=0$.
Solution:
Comparing the given equation with the general form $\mathrm{ax}^2+2 h x y+b y^2+2 g x+2 f y+c=0$ we get $\mathrm{a}=12, \mathrm{~b}=-12, \mathrm{c}=\mathrm{k}, \mathrm{f}=7 / 2, \mathrm{~g}=-1 / 2, \mathrm{~h}=7 / 2$
Here $\mathrm{a}+\mathrm{b}=0 \Rightarrow$ the given equation represents a pair of perpendicular lines
To find $\mathrm{k}$ : The condition for the given equation to represent a pair of straight lines is abc $+2 \mathrm{fgh}-\mathrm{af}^2-$ $\mathrm{bg}^2-\mathrm{ch}^2=0$

$
\begin{aligned}
& \text { (i.e) } \quad(12)(-12)(k)+8\left(\frac{7}{8}\right)\left(-\frac{1}{2}\right)\left(\frac{7}{2}\right)-12\left(\frac{49}{4}\right)+12\left(\frac{1}{4}\right)-k\left(\frac{49}{4}\right)=0 \\
& -144 k-\frac{49}{4}-147+3-\frac{49 k}{4}=0 \\
& \frac{-576 k-49 k}{4}=\frac{49+588-12}{4} \\
& -\frac{625 k}{4}=\frac{625}{4} \Rightarrow k=-1 \\
&
\end{aligned}
$
Question 13 .
For what value of $k$ does the equation $12 x^2+2 k x y+2 y^2,+11 x-5 y+2=0$ represent two straight lines. Solution:
$
12 x^2+2 k x y+2 y^2+11 x-5 y+2=0
$
Comparing this equation with the general form we get
$
a=12, b=2, c=2, f=-5 / 2, g=11 / 2, h=\frac{2 k}{2}=k
$
The condition for the given equation to represent a pair of straight lines is
$
\begin{aligned}
& a b c+2 f g h-a f^2-b g^2-c h^2=0 \\
& (\text { i.e })(12)(2)(2)+2\left(-\frac{5}{2}\right)\left(\frac{11}{2}\right)(k)-12\left(\frac{25}{4}\right)-2\left(\frac{121}{4}\right)-2\left(k^2\right)=0 \\
& 48 \frac{-55 k}{2}-75-\frac{121}{2}-2 k^2=0 \\
& \quad-27-\frac{55 k}{2}-\frac{121}{2}-2 k^2=0 \\
& (\times \text { by } 2) \quad-4 k^2-121-55 k-54=0 \\
& 4 \mathrm{k}^2+55 \mathrm{k}+175=0 \\
& 4 \mathrm{k}^2+20 \mathrm{k}+35 \mathrm{k}+175=0 \\
& 4 \mathrm{k}(\mathrm{k}+5)+35(\mathrm{k}+5)=0 \\
& (4 \mathrm{k}+35)(\mathrm{k}+5)=0 \\
& \mathrm{k}=-5 \text { or }-35 / 4
\end{aligned}
$
Question 14
Show that the equation $9 x^2-24 x y+16 y^2-12 x+16 y-12=0$ represents a pair of parallel lines. Find the distance between them.
Solution:
Comparing the given equation with $\mathrm{ax}^2+2 \mathrm{kxy}+\mathrm{by}^2=0$ we get $\mathrm{a}=9, \mathrm{~h}=-12, \mathrm{~b}=16$.
Now $\mathrm{h}^2=(-12)^2=144, \mathrm{ab}=(9)(16)=144$
$\mathrm{h}^2=\mathrm{ab} \Rightarrow$ The given equation represents a pair of parallel lines.
To find their separate equations:
$9 x^2-24 x y+16 y^2=(3 x-4 y)^2$
So, $9 x^2-24 x y+16 y^2-12 x+16 y-12=(3 x-4 y+1)(3 x-4 y+m)$
Here coefficient of $\mathrm{x} \Rightarrow 3 \mathrm{~m}+31=-12 \Rightarrow \mathrm{m}+1=-4$
coefficient of $y \Rightarrow-4 m-41=16 \Rightarrow m+1=-4$
Constant term $1 \mathrm{~m}=-12$
Now $1+\mathrm{m}=-4$ and $1 \mathrm{~m}=-12 \Rightarrow 1=-6$ and $\mathrm{m}=2$
So the separate equations are $3 x-4 y-6=0$ and $3 x-4 y+2=0$
The distance between the parallel lines is $\pm \frac{2+6}{\sqrt{3^2+4^2}}=\frac{8}{5}$ units

Question 15 .
Show that the equation $4 x^2+4 x y+y^2-6 x-3 y-4=0$ represents a pair of parallel lines. Find the distance between them.
Solution:
$
\begin{aligned}
& 4 x^2+4 x y+y^2-6 x-3 y-4=0 \\
& a=4 \\
& b=1 \\
& h=4 / 2=2 \\
& h^2-a b=2^2-(4)(1)=4-4=0
\end{aligned}
$
$\Rightarrow$ The given equation represents a pair of parallel lines.
To find the separate equations $4 x^2+4 x y+y^2=(2 x+y)^2$
So, $4 x^2+4 x y+y^2-6 x-3 y-4=(2 x+y+1)(2 x+y+m)$
Coefficient of $\mathrm{x} \Rightarrow 2 \mathrm{~m}+21=-6 \Rightarrow 1+\mathrm{m}=-3$
Coefficient of $y \Rightarrow 1+\mathrm{m}=-3$
Constant term $\Rightarrow 1 \mathrm{~m}=-4$
Now $1+\mathrm{m}=-3$ and $1 \mathrm{~m}=-4 \Rightarrow 1=-4, \mathrm{~m}=1$
So the separate equations are $2 x+y+1=0$ and $2 x+y-4=0$
The distance between the parallel lines is $\pm \frac{1+4}{\sqrt{4+1}}=\frac{5}{\sqrt{5}}=\sqrt{5}$ units
Question 16.
Prove that one of the straight lines given by $\mathrm{ax}^2+2 h x y+b y^2=0$ will bisect the angle between the coordinate axes if $(\mathrm{a}+\mathrm{b})^2=4 \mathrm{~h}^2$
Solution:
Let the slopes be 1 and $\mathrm{m}$
$\because$ One line bisects the angle between the coordinate axes $\Rightarrow \theta=45^{\circ}$
So $\tan \theta=1$

The slopes are 1 and $\mathrm{m}$
$
\begin{aligned}
& \text { Sum of the slopes }=-\frac{2 h}{b} \\
& \Rightarrow 1+m=-\frac{2 h}{b} \Rightarrow m=-\frac{2 h}{b}-1=-\frac{2 h-b}{b}
\end{aligned}
$
Product of the slopes $=\frac{a}{b} \Rightarrow(1)(m)=\frac{a}{b} \Rightarrow m=\frac{a}{b}$
$
\Rightarrow \frac{-2 h-b}{b}=\frac{a}{b} \Rightarrow a=-2 h-b \Rightarrow a+b=-2 h
$
So $(a+b)^2=(-2 h)^2=4 h^2$
(i.e) $(a+b)^2=4 h^2$
Question 17.
If the pair of straight lines $x^2-2 k x y-y^2=0$ bisect the angle between the pair of straight lines $x^2-21 x y$ $-y^2=0$, show that the later pair also bisects the angle between the former.
Solution:
Given that $x^2-2 k x y-y^2=0$
Bisect the angle between the lines $x^2-11 x y-y^2=0$

Equation of the angle bisector of line (2) is $\left(\frac{x^2-y^2}{a+b}=\frac{x y}{h}\right)$
$
\begin{aligned}
& \text { (i.e) } \frac{x^2-y^2}{2}=\frac{x y}{-l} \\
& \Rightarrow x^2-y^2=\frac{-2 x y}{l} \Rightarrow x^2-\frac{2}{l} x y-y^2=0
\end{aligned}
$
Equations (2) and (3) represents the same
$
\Rightarrow \frac{1}{1}=\frac{-2 k}{2 / l}=\frac{-1}{-1} \Rightarrow 1=\frac{-2 l k}{2}=1 \Rightarrow l k=-1
$
To prove that $x^2-2 l x y-y^2=0$ is the angle bisector of $x^2-2 k x y-y^2=0$ Equation of the bisector of the lines $x^2-2 k x y-y^2=0$ is
$
\left(\frac{x^2-y^2}{a+b}=\frac{x y}{h}\right) \Rightarrow \frac{x^2-y^2}{2}=\frac{x y}{-2 k} \Rightarrow x^2-y^2=\frac{2 x y}{-k}
$
(i.e) $x^2-y^2=-\frac{2 x y}{k}$
(i.e) $x^2+\frac{2 x y}{k}-y^2=0$
Using $l k=-1$ we get $k=-1 / l$.
So, $x^2+\frac{2 x y}{k}-y^2=0$ becomes $x^2-2 l x y-y^2=0$ is the bisector of the pair of straight line $x^2-2 k x y-y^2=0$
Question 18.
Prove that the straight lines joining the origin to the points of intersection of $3 x^2+5 x y-3 y^2+2 x+3 y=$ 0 and $3 x-2 y-1=0$ are at right angles.
Solution:
Homogenizing the given equations $3 x^2+5 x y-3 y^2+2 x+3 y=0$ and $3 x-2 y-1=0$ (i.e) $3 x-2 y=1$.
We get $\left(3 x^2+5 x y-3 y^2\right)+(2 x+3 y)(1)=0$
(i.e) $\left(3 x^2+5 x y-3 y^2\right)+(2 x+3 y)(3 x-2 y)=0$
$3 x^2+5 x y-3 y^2+6 x^2-4 x y+9 x y-6 y^2=0$
$9 x^2+10 x y-9 y^2=0$
Coefficient of $x^2+$ coefficient of $y^2=9-9=0$
$\Rightarrow$ The pair of straight lines are at right angles.