Find the equation of the locus of points equi | vedclass

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

Question

(B) The locus of points equidistant from two lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ is given by the angle bisectors: $\frac{a_1x +

Vedclass · Accepted Answer

$99x + 77y - 133 = 0$

Vedclass · Answer

(B) The locus of points equidistant from two lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ is given by the angle bisectors: $\frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}}$.Substituting the given lines: $\frac{3x + 4y - 11}{\sqrt{3^2 + 4^2}} = \pm \frac{12x + 5y + 2}{\sqrt{12^2 + 5^2}}$.$\frac{3x + 4y - 11}{5} = \pm \frac{12x + 5y + 2}{13}$.Case $1$: $13(3x + 4y - 11) = 5(12x + 5y + 2) \implies 39x + 52y - 143 = 60x + 25y + 10 \implies 21x - 27y + 153 = 0 \implies 7x - 9y + 51 = 0$.Case $2$: $13(3x + 4y - 11) = -5(12x + 5y + 2) \implies 39x + 52y - 143 = -60x - 25y - 10 \implies 99x + 77y - 133 = 0$.Comparing with the options,the correct equation is $99x + 77y - 133 = 0$.

Find the equation of the locus of points equidistant from the lines $3x + 4y - 11 = 0$ and $12x + 5y + 2 = 0$.

Similar Questions

The equations of the angle bisectors between the lines $3x - 4y + 7 = 0$ and $12x - 5y - 8 = 0$ are:

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

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

Let $P \equiv (-5, 0)$,$Q \equiv (0, 0)$,and $R \equiv (2, 2\sqrt{3})$ 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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module