The equation of pair of lines $y=\mathrm{p} x$ and $y=\mathrm{q} x$ can be written as $(y-\mathrm{p}…
The equation of pair of lines $y=\mathrm{p} x$ and $y=\mathrm{q} x$ can be written as $(y-\mathrm{p} x)(y-\mathrm{q} x)=0$. Then the equation of the pair of the angle bisectors of the lines $x^2-4 x y-5 y^2=0$ is
$x^2-3 x y+y^2=0$
$x^2+4 x y-y^2=0$
$x^2-3 x y-y^2=0$
$x^2+3 x y-y^2=0$
Solution
Equation of angle bisector of two lines whose general equation is $a x^2+2 \mathrm{~h} x y+b y^2=0$ is
$\frac{x^2-y^2}{a-b}=\frac{x y}{h}$
$\therefore \quad$ Comparing given equation $x^2-4 x y-5 y^2=0$ with $\mathrm{a} x^2+2 \mathrm{~h} x y+\mathrm{b} y^2=0$, we get
$a=1, b=-5, h=-2$
$\therefore \quad$ Equation of angle bisectors is
$\begin{aligned}
& \frac{x^2-y^2}{1-(-5)}=\frac{x y}{-2} \\
& \Rightarrow x^2-y^2=-3 x y \\
& \Rightarrow x^2+3 x y-y^2=0
\end{aligned}$