Find the value of k for which the points (–5, 1), (1, k) and (4, –2) are collinear.

Given:

A (say) = (–5, 1)

B (say) = (1, k)

C (say) = (4, –2)

For A, B and C to be collinear they would lie in same line and therefore cannot make triangle.

i.e.,

Area of ∆ABC should be 0.

We know that,

Area of ∆ABC = 1/2[x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]

⇒ Area of ∆ABC = 1/2[(-5)(k – (-2)) + 1((-2) – 1) + 4(1 – k)] = 0

⇒ 1/2[(-5)(k + 2) + 1(-3) + 4(1 – k)] = 0

⇒ 1/2[-5k – 10 – 3 + 4 – 4k] = 0

⇒ -5k – 10 – 3 + 4 – 4k = 0

⇒ -9k – 9 = 0

⇒ -9(k + 1) = 0

⇒ k + 1 = 0

⇒ k = -1

Hence, k = -1