Question

One of the bisector of the angle between the lines
$a{(x-1)}^2+2h(x-1)(y-1)+b{(y-2)}^2=0$ is $x+2y-5=0$. The other bisector is

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

Answer 1

Answer: (A)

$2x-y=0$

We know that bisectors are perpendicular to each other.

Given one bisector is $x+2y-5=0$

Therefore the other bisector is of the form $2x-y+\lambda =0$

By the given alternatives, the one which matches is $2x-y=0$

Therefore the other bisector is $2x-y=0$

Hence $\left(A\right)$ is correct.