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Every Point on the Bisector of an Angle Is Equidistant from the Sides of the Angle.

If one of the...

Question

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 $x y = 0$ then $m$ is -

- \frac{1}{2}

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- 2

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Solution

The correct option is C $1$
$\begin{matrix} H e r e . E q u a t i o n o f t h e a n g l e b e t w e e n t h e l i n e s x y = 0 i s y = \pm x p u t y = \pm x i n m y^{2} + (1 - m^{2}) x y - m x^{2} = 0 w e g e t m = \pm 1 H e n c e, t h e o p t i o n C i s t h e c o r r e c t a n s w e r . \end{matrix}$

Suggest Corrections

Similar questions

m y^{2} + (1 - m^{2}) x y - m x^{2} = 0

m y^{2} + (1 - m^{2}) x y - m x^{2} = 0

x y = 0,

m

{m y}^{2} + ({1 - m}^{2}) x y - {m x}^{2} = 0

The equation of the bisectors of angle between the lines represented by equation $(y - m x)^{2} = (x + m y)^{2}$ is

x^{2} - 2 m x y - y^{2} = 0

x^{2} - 2 n x y - y^{2} = 0

1 + m n = 0

a x^{2} + 2 h x y + b y^{2} = 0

\frac{x^{2} - y^{2}}{a - b} = \frac{x y}{h} .

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