The given points A(−1, −4), B(b, c) and C(5, −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

∴ −1[c − (− 1)]+b[− 1 − ( − 4)] + 5( − 4 − c) = 0

∴ − c − 1 + 3b − 20 − 5c = 0

3b − 6c = 21

∴b − 2c = 7 …(1)

Also it is given that 2b + c = 4 …(2)

Solving 1 and 2 simultaneously, we get,

2(7 + 2c) + c = 4

14 + 4c + c = 4

5c = − 10

c = − 2

∴ b = 3

Hence, value of b and c are 3 and -2 respectively