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

The coordinate axes are represented by the pair of lines $$xy = 0$$, i.e. $$x = 0$$ and $$y = 0$$. The bisectors of the angle between these two perpendicular lines are well known:

• the line $$y = x$$ (the internal bisector),
• the line $$y = -x$$ (the external bisector).

Hence, if one of the factors of the second-degree curve
$$my^2 + (1 - m^2)xy - mx^2 = 0$$
is a bisector, then either $$y = x$$ or $$y = -x$$ must satisfy the equation identically.

Let $$y = x$$. Substitute $$y = x$$ (so $$y^2 = x^2$$ and $$xy = x^2$$) into the given expression:

$$m\,x^2 + (1 - m^2)\,x^2 - m\,x^2 = 0$$
$$\Longrightarrow \bigl[m + (1 - m^2) - m\bigr]x^2 = 0$$
$$\Longrightarrow (1 - m^2)x^2 = 0.$$

Since $$x \neq 0$$ for a non-trivial line, we need
$$1 - m^2 = 0 \; \Longrightarrow \; m^2 = 1.$$

Let $$y = -x$$. Substitute $$y = -x$$ (so $$y^2 = x^2$$ and $$xy = -x^2$$) into the curve equation:

$$m\,x^2 + (1 - m^2)(-x^2) - m\,x^2 = 0$$
$$\Longrightarrow \bigl[m - (1 - m^2) - m\bigr]x^2 = 0$$
$$\Longrightarrow (m^2 - 1)x^2 = 0.$$

Again, for a genuine line $$x \neq 0$$, so we obtain
$$m^2 - 1 = 0 \; \Longrightarrow \; m^2 = 1.$$

Both cases lead to the single condition $$m^2 = 1$$, i.e. $$m = \pm1$$. Among the answer choices given, only $$m = 1$$ appears.

Therefore, the required value is
Option C which is: 1.