Question

Let $A(3,2,0),B(5,3,2),C(-9,6,-3)$ be three points forming a triangle. $AD$, the bisectors of $\angle BAC$,meets $BC$ an $D$. Find the coordinates of the point $D$.

Answer 1

We have,

Given points are

$A(x_1,y_1,z_1)=(3,2,0)$

$B(x_2,y_2,z_2)=(5,3,2)$

$C(x_3,y_3,z_3)=(-9,6,-3)$

So,

Point D be the midpoint of AC

Then,

Coordinates of point

$D(x_4,y_4,z_4)=(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2},\frac{z_2+z_3}{2})$

$D(x_4,y_4,z_4)=(\frac{5-9}{2},\frac{3+6}{2},\frac{2-3}{2})$

$D(x_4,y_4,z_4)=(\frac{-4}{2},\frac{9}{2},\frac{-1}{2})$

$D(x_4,y_4,z_4)=(-2,\frac{9}{2},\frac{-1}{2})$

Hence, this is the answer.