The vertices of a triangle ABC are A(3, 2, 0), B(5, 3, 2) and C(-9, 6, -3). The bisector AD of ∠ A meets BC at D, find the fourth vertex D.
The given co-ordinates: A(3, 2, 0), B(5, 3, 2) and C(-9, 6, -3)
Now, AB = 3
Also, AC
Now, we have,
By the property of internal angle bisector,
Therefore,
Applying the section formula, we get,
D(x, y, z)
D(x, y, z)
Answer.