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