# A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle ∠BAC meets BC.

220 views

closed

A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle ∠BAC meets BC.

+1 vote
by (32.4k points)
selected

Given: The vertices of the triangle are A(1, 2, 3), B(0, 4, 1) and C(-1, -1, -3)

To find: the coordinates of D

Formula used:

Distance Formula:

The distance between any two points (a, b, c) and (m, n, o) is given by Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c). The coordinates of C is given by,

$\Big(\frac{nx+ma}{m+n},\frac{ny+mb}{m+n},\frac{nz+mc}{m+n}\Big)$

We know angle bisector divides opposite side in the ratio of the other two sides.

As AD is angle bisector of A and meets BC at D

⇒ BD : DC = AB : BC Distance between A(1, 2, 3) and B(0, 4, 1) is AB, = $\sqrt{9}$

= 3

Distance between A(1, 2, 3) and C(-1, -1, -3) is AC, AB : AC = 3:7

⇒ BD: DC = 3:7

Therefore, m = 3 and n = 7

B(0, 4, 1) and C(-1, -1, -3)

Coordinates of D using section formula: Hence, Coordinates of D are $\Big(\frac{-3}{10}, \frac{5}{2}, \frac{-1}{5}\Big)$