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Q. 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

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Solution:

Equation of bisectors of lines, $xy = 0$ are $ y = \pm \, x $
$\therefore $ Put $y = \pm x$ in the given equation
$my^2 + (1 - m^2)xy - mx^2 = 0$
$\therefore \, mx^2 + (1 - m^2)x^2 - mx^2 = 0 $
$\Rightarrow \, 1 - m^2 = 0 \, \Rightarrow \, m = \pm 1 $

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