Question

The combined equation of the two lines $a x+b y+c=0$ and $a^{\prime} x+b^{\prime} y+c^{\prime}=0$ can be written as $(a x+b y+c)\left(a^{\prime} x+b^{\prime} y+c^{\prime}\right)=0$.
The equation of the angle bisectors of the lines represented by the equation $2 x^2+x y-3 y^2=0$ is

• A. $3 x^2+5 x y+2 y^2=0$
• B. $x^2-y^2-10 x y=0$
• C. $3 x^2+x y-2 y^2=0$
• D. $x^2-y^2+10 x y=0$

Answer 1

Answer: (B)

Equation of the pair of angle bisector for the homogenous equation $a x^2+2 h x y+b y^2=0$ is given as
$\frac{x^2-y^2}{a-b}=\frac{x y}{h}$
Here $a=2, h=1 / 2 \& b=-3$
Equation will become
$ \frac{x^2-y^2}{2-(-3)}=\frac{x y}{1 / 2} $
$ x ^2-y^2=10 x y $
$ x^2-y^2-10 x y=0$