Question

If one of the lines of $m{y}^2+(1-m^2)xy-m{x}^2=0$ is a bisector of the angle between the lines $xy=0$, then $m$ can be

• A. $1$
• B. $2$
• C. $-1$
• D. $-\frac{1}{2}$

Answer 1

Answer: (A), (C)

$-1$

$m{y}^2+(1-m^2)xy-m{x}^2=0\Rightarrow m{y}^2+xy-(m^{2}xy+m{x}^2)=0\Rightarrow (x+my)(y-mx)=0$
So lines will be $x+my=0;y-mx=0\cdots \left(1\right)$
Now both the lines are perpendicular to each other
and $xy=0$
$y=0;x=0$ i.e. coordinate axes
hence angle bisector equation will be
$y=x$ and $x+y=0\cdots \left(2\right)$
compairing $\left(1\right)$ and $\left(2\right)$
we get the value of $m=\pm 1$

Alternate Solution :
the pair of lines represented by $m{y}^2+(1-m^2)xy-m{x}^2=0$ are perpendicular
and bisector of the pair of lines $xy=0$ are $x=y,x+y=0$
substituting $y=\pm x$ in the $m{y}^2+(1-m^2)xy-m{x}^2=0$
we get $m^2-1=0$
$\Rightarrow m=\pm 1$