How to prove three points collinear by Section Formula - Maths - Coordinate Geometry

Question

How to prove three points collinear by Section Formula

Ankush Jain · Accepted Answer

Let us understand this concept by taking three points A(2, 5), B(4, 6) and C (8, 8).

Now, we have to prove that the points A(2, 5), B(4, 6) and C (8, 8) are Collinear.

Let B divides AC in the ratio k : 1

Then the coordinates of B will be

$(\frac{8 k + 2}{k + 1}, \frac{8 k + 5}{k + 1})$

But the coordinates of B are (4, 6)

Comparing we get

$\frac{8 k + 2}{k + 1} = 4 ��and �� \frac{8 k + 5}{k + 1} = 6$

⇒8k + 2 = 4k + 4 are 8k + 5 = 6k + 6

⇒8k – 4k = 4 – 2 are 8k – 6k = 6 – 5

⇒ 4k = 2 are 2k = 1

$\Rightarrow k = \frac{2}{4} = \frac{1}{2} �and � k = \frac{1}{2}$

Since the value of k is same in both x and y coordinates

∴ Points A, B and C are collinear.

Hope you get it!!