Question -
Answer -
Given:
The points A (3, 2,-4), B (9, 8, -10) and C (5, 4, -6)
By using the sectionformula,
Let C divides AB inratio k: 1
Three points arecollinear if the value of k is the same for x, y and z coordinates.
Then, m = k and n = 1
A(3, 2, -4), B(9, 8,-10) and C(5, 4, -6)
Coordinates ofC are:
On comparing we get,
[9k +3] / [k + 1] = 5
9k + 3 = 5(k + 1)
9k + 3 = 5k + 5
9k – 5k = 5 – 3
4k = 2
k = 2/4
= ½
[8k +2] / [k + 1] = 4
8k + 2 = 4(k + 1)
8k + 2 = 4k + 4
8k – 4k = 4 – 2
4k = 2
k = 2/4
= ½
[-10k– 4] / [k + 1] = -6
-10k – 4 = -6(k + 1)
-10k – 4 = -6k – 6
-10k + 6k = 4 – 6
-4k = -2
k = -2/-4
= ½
The value of k is thesame in all three cases.
So, A, B and Care collinear [as, k = ½]
∴We can say that, Cdivides AB externally in ratio 1:2