If one of the lines of my2 + (1- m2) xy - mx2= 0 is a bisector of the angle between the lines xy = 0, then m is

Question

If one of the lines of my2 + (1- m2) xy - mx2= 0 is a bisector of the angle between the lines xy = 0, then m is (A) 1 (B) 2 (C) -1/2 (D) -2

Tardigrade Admin · Accepted Answer

Equation of bisectors of lines, xy = 0 are y = ± x ∴ Put y = ± x in the given equation my2 + (1 - m2)xy - mx2 = 0 ∴ mx2 + (1 - m2)x2 - mx2 = 0 ⇒ 1 - m2 = 0 ⇒ m = ± 1

Q. If one of the lines of $m y^{2} + (1 - m^{2}) x y - m x^{2} = 0$ is a bisector of the angle between the lines xy = 0, then m is

1

2

-1/2

-2

Equation of bisectors of lines, $x y = 0$ are $y = \pm x$
$∴$ Put $y = \pm x$ in the given equation
$m y^{2} + (1 - m^{2}) x y - m x^{2} = 0$
$∴ m x^{2} + (1 - m^{2}) x^{2} - m x^{2} = 0$
$\Rightarrow 1 - m^{2} = 0 \Rightarrow m = \pm 1$

Q. If one of the lines of my2+(1−m2)xy−mx2=0 is a bisector of the angle between the lines xy = 0, then m is

1

2

-1/2

-2

Equation of bisectors of lines, xy=0 are y=±x ∴ Put y=±x in the given equation my2+(1−m2)xy−mx2=0 ∴mx2+(1−m2)x2−mx2=0 ⇒1−m2=0⇒m=±1

Q. If one of the lines of $m y^{2} + (1 - m^{2}) x y - m x^{2} = 0$ is a bisector of the angle between the lines xy = 0, then m is

Equation of bisectors of lines, $x y = 0$ are $y = \pm x$
$∴$ Put $y = \pm x$ in the given equation
$m y^{2} + (1 - m^{2}) x y - m x^{2} = 0$
$∴ m x^{2} + (1 - m^{2}) x^{2} - m x^{2} = 0$
$\Rightarrow 1 - m^{2} = 0 \Rightarrow m = \pm 1$