Question

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

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

Answer 1

Answer: (A), (D)

The correct options are
A $1$
D $-1$

Given pair of lines is
${my}^2+(1-m^2)xy-{mx}^2=0\Rightarrow m{\left(\frac{y}{x}\right)}^2+{(1-m)}^2\frac{y}{x}-m=0$ ...(1)

Lines $xy=0$ are $x=0$ and $y=0.$
i.e the axes bisector of angle between the axes are $y=x$ and $y=-x$ $\Rightarrow \frac{y}{x}=1$ or $\frac{y}{x}=-1$

If $\frac{y}{x}=1$ is represented by (1), then
$m{\left(1\right)}^2+(1-m^2)(-1)-m=0\Rightarrow 1-m^2=0\Rightarrow m=\pm 1$

If $\frac{y}{x}=-1$ is represented by (1), then
$m{(-1)}^2+(1-m^2)(-1)-m=0\Rightarrow 1-m^2=0\Rightarrow m=\pm 1$
In both cases either $m=1$ or $m=-1.$