Let the equation of the pair of lines, y=px and y=qx can be written as (y-px)(y-qx)=0.

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 line $x^2-4xy-5y^2=0$ is:

Accepted Answer

For $ax^2+2hxy+by^2=0$, bisector pair: $\dfrac{x^2-y^2}{a-b}=\dfrac{xy}h$.

Here $a=1,\ h=-2,\ b=-5$:
$$\frac{x^2-y^2}{1-(-5)}=\frac{xy}{-2} \implies \frac{x^2-y^2}6=-\frac{xy}2 \implies x^2-y^2=-3xy$$

$x^2+3xy-y^2=0$

Solutions · 1