If the points A ( - 2,1), B (a, b) and C (4, - 1) are collinear and a – b = 1, find the values of a and b.

Given: Points A ( - 2,1), B (a, b) and C (4, - 1) are collinear.

a – b = 1

To find: The values of a and b.

Formula Used:

Area of the triangle having vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃)
= 1/2 |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|

Explanation:

The given points A (−2, 1), B (a, b) and C (4, −1) are collinear.

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃)
= 1/2 |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|

Given that area of ∆ABC = 0

∴ −2[b − (− 1)] + a (− 1 – 1) + 4 ( 1 –b ) = 0

- 2b – 2 − 2a + 4 − 4b = 0

− 2a − 6b = − 2

a + 3b = 1 … (1)

Also, it is given that a – b = 1

a = b + 1 …. (2)

Put this value in (1) to get,

b + 1 + 3b = 1

4b = 0

∴ b = 0

Put this value in (2) to get,

a = 0 + 1

∴ a = 1

Hence, the values of a and b are 1 and 0.