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 $m$ is/are

The correct option is C

$\pm 1$

Explanation for the correct option

Step 1: Information required for the solution

As we know that $xy=0$, this implies that either $x$ can be equal to zero or $y$ can be equal to zero.

From this, we can construct two equations which are,

$$\begin{gathered} x+y=0 \\ x-y=0 \end{gathered}$$

Step 2: Simplification of the equation $\mathit{m}{\mathit{y}}^{\mathbf{2}}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{-}{\mathit{m}}^{\mathbf{2}}\mathbf{)}\mathit{x}\mathit{y}\mathbf{-}\mathit{m}{\mathit{x}}^{\mathbf{2}}\mathbf{=}\mathbf{0}$,

The equation can be simplified as,

$\begin{array}{rcl}m{y}^2+(1-m^2)xy-m{x}^2& =& 0\\ & \Rightarrow & m{y}^2+xy-m^{2}xy-m{x}^2=0\\ & \Rightarrow & y\left(my+x\right)-mx\left(my+x\right)=0\\ & \Rightarrow & \left(y-mx\right)\left(my+x\right)=0\end{array}$

From this, we get $y=mx\text{and}my=-x$

Step 3: Calculation of the value of $\mathit{m}$.

Now, we know that $x=-y\text{or}x=y$ then apply this to the equations $y=mx\text{and}my=-x$.

So, we conclude that $m$ can be equal to $1\text{or}-1$.

Hence, the correct option is (C).