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

The Correct Option is B

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

Answer 2

Step 1: Equation of the pair of angle bisectors for the homogeneous equation \[ ax^2 + 2hxy + by^2 = 0 \] is given by: \[ \frac{x^2 - y^2}{a - b} = \frac{xy}{h} \] 

Step 2: Here, \[ a = 2, \quad h = \frac{1}{2}, \quad b = -3 \] Thus, equation becomes: \[ \frac{x^2 - y^2}{2 - (-3)} = \frac{xy}{\frac{1}{2}} \] \[ x^2 - y^2 - 10xy = 0 \]