Question

The vertices of a triangle are A (-1, -7), B (5, 1) and C (1, 4). Then what is the equation of the bisector of the angle ∠ABC?

Answer 1

Answer:

hope so it helps you please thank my answers

Answer 2

Answer:

The vertices of the triangle are given to be A (-1, -7), B (5, 1) and C (1, 4).

Hence, the equation of the line AB is (y-1) = [1-(-7)] / [5-(-1)] (x-5)

Hence, (y-1) = 8/6 (x-5)

This gives (y-1) = 4/3 (x-5)

We have 3y - 3 = 4 x – 20.

Hence, the equation becomes 3y-4x+17 = 0

Equation of the line BC is (y-4) = (4-1)/(1-5) (x-1)

Hence, (y-4) = -3/4(x-1)

This yields, 4y-16 = -3x+3

Hence, 3x+4y-19 = 0

Again, the equation of the bisectors of angles between two given lines AB and BC are

(3y-4x+17)/√32+42 = ± (4y+3x-19) / √42 + 32

This gives (3y-4x+17) = ± (4y+3x-19)

Hence, we have two relations

(3y-4x+17) = (4y+3x-19)

and (3y-4x+17) = - (4y+3x-19)

This gives y +7x = 36 and 7y-x = 2

Out of these two, the equation of the bisector of ∠ABC is 7y = x + 2.