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)

= |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 …(2)

Solving 1 and 2 simultaneously,

B + 1 + 3b = 1

4b = 0

∴ b = 0

∴ a =1

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