A (2, 5), B (–1, 2) and C (5, 8) are the vertices of a triangle ABC, ‘M’ is a point on AB such that AM: MB = 1: 2. Find the co-ordinates of ‘M’. Hence find the equation of the line passing through the points C and M.

Given:

A(2, 5) B(−1, 2) C(5, 8)

AM : MB = 1 : 2

Using the section formula:

M = ((1×(−1) + 2×2)/(1+2), (1×2 + 2×5)/(1+2))

= ((−1 + 4)/3, (2 + 10)/3)

= (1, 4)

Now find the slope of CM:

m = (8 − 4)/(5 − 1)

= 4/4

= 1

Equation of the line through M(1,4):

y − 4 = 1(x − 1)

y − 4 = x − 1

x − y + 3 = 0

Answer:

M = (1, 4) Equation of CM: x − y + 3 = 0.

The point M divides AB internally in the ratio 1:2, so the section formula gives M(1,4). The line CM then has slope 1, and the point-slope form gives its equation.