Question

If the pairs of lines $x^2-2nxy-y^2=0$ and $x^2-2mxy-y^2=0$ are such that one of them represents the bisectors of the angles between the other, then

• A. $\frac{1}{n}+\frac{1}{m}=0$
• B. $\frac{1}{n}-\frac{1}{m}=0$
• C. $nm-1=0$
• D. $nm+1=0$

Answer 1

The correct option is D $nm+1=0$

Equation of bisectors of the angles between the lines is given by;

$x^2-2mxy-y^2=0$

$\frac{x^2-y^2}{1-(-1)}=\frac{xy}{-m}$

$x^2+\left(\frac{2}{m}\right)xy-y^2=\mathrm{0.......................}\left(1\right)$

Given that $x^2-2nxy-y^2=\mathrm{0...........}\left(2\right)$

Comparing (1) and (2) we get

$\frac{2}{m}=-2n$

$-mn=1$

$nm+1=0$