Question

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

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

Answer 1

The correct option is B -2

In pair of line given by $xy=0$

we have lines, $x=0$ and $y=0$

Given prair of lines $\to m{y}^2+(1-m^2)xy-m{x}^2=0$

$\Rightarrow m{y}^2+xy-m^{2}xy-m{x}^2=0$

$\Rightarrow m^{2}xy+m(x^2-y^2)-xy=0$

It is a quadratic in m

so product of roots of this quadratic $=-\frac{xy}{xy}$

$=-1$

so lines cantained in given equation

are perpendicular to each other

$m{y}^2+xy-m^{2}xy-m{x}^2=0$

$\Rightarrow y(my+x)-mx(my+x)=0$

$\Rightarrow (my+x)(y-mx)=0$

$\therefore$. either $y=-\frac{x}{m}$ or $y=mx$

We need bisectars of $x=0$ and $y=0$ and we know that it can by either $x=y$ or $x=-y$

so comparing $x=y$ with $y=-\frac{x}{m}$ and $y=mx$

we get $m=-1$ or $m=1$

camparing $x=-y$ with $y=-\frac{x}{m}$ and $y=mx$

we get $m=-1$ or $m=1$

so Answer:option $\left(c\right)$