The family of lines,forming an isosceles tria | vedclass

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

Question

(C) The family of lines forming an isosceles triangle with two given lines must be parallel to the angle bisector of the angle between the two lines.F

Vedclass · Accepted Answer

$9x - 7y + c = 0$

Vedclass · Answer

(C) The family of lines forming an isosceles triangle with two given lines must be parallel to the angle bisector of the angle between the two lines.First,we find the angle bisectors of the lines $L_1: 3x - 4y - 2 = 0$ and $L_2: 12x - 5y + 6 = 0$.The equation of the bisectors is given by $\frac{3x - 4y - 2}{\sqrt{3^2 + (-4)^2}} = \pm \frac{12x - 5y + 6}{\sqrt{12^2 + (-5)^2}}$.$\frac{3x - 4y - 2}{5} = \pm \frac{12x - 5y + 6}{13}$.Taking the negative sign (to get the bisector of the angle containing the origin or specific orientation):$13(3x - 4y - 2) = -5(12x - 5y + 6)$$39x - 52y - 26 = -60x + 25y - 30$$99x - 77y + 4 = 0$.Dividing by $11$,we get $9x - 7y + \frac{4}{11} = 0$.Thus,the family of lines parallel to this bisector is $9x - 7y + c = 0$.

The family of lines,forming an isosceles triangle with the lines $3x - 4y - 2 = 0$ and $12x - 5y + 6 = 0$,is

Similar Questions

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 the angle bisectors between the lines $3x + 4y - 7 = 0$ and $12x + 5y + 17 = 0$ is:

The bisector of the acute angle formed between the lines $4x - 3y + 7 = 0$ and $3x - 4y + 14 = 0$ has the equation

If the vertices of a triangle are $A(-1, -7)$,$B(5, 1)$,and $C(1, 4)$,then the equation of the angle bisector of $\angle ABC$ is:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module