Question

Find the value of k for which the points A(−1, 3), B(2, k) and C(5, −1) are collinear.

Answer 1

Let A(−1, 3), B(2, k) and C(5, − 1) be the given points. Then,
(x₁ = − 1, y₁ = 3), (x₂ = 2 , y₂ = k) and (x₃ = 5, y₃ = − 1)
Since the given points are collinear, the area of the triangle formed by them must be 0.
⇒$\frac{1}{2}$[x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)] = 0
⇒ − 1(k + 1) + 2( − 1 − 3) + 5(3 − k) = 0
⇒ − 1(k + 1) + 2( − 4) + 5(3 − k) = 0
⇒ − k − 1 − 8 + 15 − 5k = 0
⇒ − 6k = − 6
⇒ 6k = 6 ⇒ k = 1
Hence, the required value of k is 1.