If one of the lines my^2 + (1 - m^2)xy - mx^2 | vedclass

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(A) The given pair of lines is $my^2 + (1 - m^2)xy - mx^2 = 0$.This can be rewritten as $-mx^2 + (1 - m^2)xy + my^2 = 0$,or $mx^2 - (1 - m^2)xy - my^2

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$1$

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(A) The given pair of lines is $my^2 + (1 - m^2)xy - mx^2 = 0$.This can be rewritten as $-mx^2 + (1 - m^2)xy + my^2 = 0$,or $mx^2 - (1 - m^2)xy - my^2 = 0$.The lines $xy = 0$ represent the coordinate axes,$x = 0$ and $y = 0$.The angle bisectors of the lines $x = 0$ and $y = 0$ are given by $x^2 - y^2 = 0$,which implies $x = y$ and $x = -y$.If one of the lines $my^2 + (1 - m^2)xy - mx^2 = 0$ is a bisector,it must be either $x - y = 0$ or $x + y = 0$.For $x - y = 0$,the equation is $x^2 - xy - xy + y^2 = 0$ (not matching) or by substituting $y = x$ into the original equation:$m(x^2) + (1 - m^2)x^2 - mx^2 = 0 \Rightarrow (1 - m^2)x^2 = 0$,which implies $m^2 = 1$,so $m = \pm 1$.However,if $m = 1$,the equation becomes $y^2 - x^2 = 0$,i.e.,$(y - x)(y + x) = 0$,which are the bisectors of $xy = 0$.If $m = -1$,the equation becomes $-y^2 + 2xy + x^2 = 0$,which are not the bisectors.Thus,$m = \pm 1$. Given the options,$m = 1$ is the correct choice.

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

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