Given:
The points A (2, 3, 4), B (-1, 2, -3) and C (-4, 1, -10)
By using the section formula,
Let C divides AB in ratio k: 1
Three points are collinear if the value of k is the same for x, y and z coordinates.
So, m = k and n = 1
A(2, 3, 4), B(-1, 2, -3) and C(-4, 1, -10)
Coordinates of C are:
On comparing we get,
[-k + 2] / [k + 1] = -4
-k + 2 = -4(k + 1)
-k + 2 = -4k – 4
4k – k = – 2 – 4
3k = -6
k = -6/3
= -2
[2k + 3] / [k + 1] = 1
2k + 3 = k + 1
2k – k = 1 – 3
k = – 2
[-3k + 4] / [k + 1] = -10
-3k + 4 = -10(k + 1)
-3k + 4 = -10k – 10
-3k + 10k = -10 – 4
7k = -14
k = -14/7
= -2
The value of k is the same in all three cases.
So, A, B and C are collinear [as k = -2]
∴ We can say that, C divides AB externally in ratio 2:1
