Question

Let the equation of the pair of lines, $y=px$ and $y=qx$, can be written as $(y-px)(y-qx)=0$. Then the equation of the pair of the angle bisectors of the lines $x^2-4xy-5y^2=0$ is

• A. $x^2-3xy+y^2=0$
• B. $x^2+3xy-y^2=0$
• C. $x^2-3xy-y^2=0$
• D. $x^2+4xy-y^2=0$

Answer 1

The correct option is B $x^2+3xy-y^2=0$

Equation of angle bisector of homogeneous equation of pair of staright lines $a{x}^2+2hxy+b{y}^2=0$ is
$\frac{x^2-y^2}{a-b}=\frac{xy}{h}$
For $x^2-4xy-5y^2=0$
$\Rightarrow a=1,h=-2,b=-5$
So, combined equation of angle bisector is
$\frac{x^2-y^2}{6}=\frac{xy}{-2}$
$\therefore x^2+3xy-y^2=0$