### Straight Lines Question 10

#### Question 10 - 01 February - Shift 1

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 $(a x+b y+c)(a^{\prime} x+b^{\prime} y+c^{\prime})=0$.The equation of the angle bisectors of the lines represented by the equation $2 x^{2}+x y-3 y^{2}=0$ is

(1) $3 x^{2}+5 x y+2 y^{2}=0$

(2) $x^{2}-y^{2}+10 x y=0$

(3) $3 x^{2}+x y-2 y^{2}=0$

(4) $x^{2}-y^{2}-10 x y=0$

## Show Answer

#### Answer: (4)

#### Solution:

#### Formula: Angle bisectors between the lines represented by homogeneous equation

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

$\frac{x^{2}-y^{2}}{2-(-3)}=\frac{x y}{1 / 2}$

$x^{2}-y^{2}=10 xy$

$x^{2}-y^{2}-10 x y=0$