The equation of bisectors of the angles betwe | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given equation $|x| = |y|$ represents the pair of lines $x = y$ and $x = -y$,which can be written as $x - y = 0$ and $x + y = 0$.To find the b

Vedclass · Accepted Answer

$y = 0$ and $x = 0$

Vedclass · Answer

(C) The given equation $|x| = |y|$ represents the pair of lines $x = y$ and $x = -y$,which can be written as $x - y = 0$ and $x + y = 0$.To find the bisectors of the angles between these lines,we use the formula $\frac{x - y}{\sqrt{1^2 + (-1)^2}} = \pm \frac{x + y}{\sqrt{1^2 + 1^2}}$.This simplifies to $\frac{x - y}{\sqrt{2}} = \pm \frac{x + y}{\sqrt{2}}$,which implies $x - y = \pm (x + y)$.Taking the positive sign: $x - y = x + y \implies 2y = 0 \implies y = 0$.Taking the negative sign: $x - y = -(x + y) \implies x - y = -x - y \implies 2x = 0 \implies x = 0$.Thus,the equations of the bisectors are $x = 0$ and $y = 0$.

The equation of bisectors of the angles between the lines $|x| = |y|$ are

Similar Questions

The equation of the bisector of the acute angle between the lines $2x - y + 4 = 0$ and $x - 2y - 1 = 0$ is

The equation of the bisector of the acute angle between the lines $3x - 4y + 7 = 0$ and $12x + 5y - 2 = 0$ is

Let $P \equiv (-3, 0)$,$Q \equiv (0, 0)$,and $R \equiv (3, 3\sqrt{3})$ be three points. Then the equation of the bisector of the angle $\angle PQR$ is

The equation of the perpendicular bisector of the line segment joining the points $(1, 2)$ and $(-2, 0)$ is:

The set of values of $\alpha$ for which the angle bisector of the lines $(\alpha + 1)x + 2y + 5 = 0$ and $4x + \alpha y - 3 = 0$ containing the origin is also the obtuse angle bisector,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module