Question

If the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k − 1, 5k) are collinear, then find the value of k

Answer 1

Since the given points are collinear, the area of the triangle formed by them must be 0.

`=> 1/2 |x_1(y_2-y_3) + x_2(y_3 - y_1) +x_3(y_1 - y_2)| = 0`

Here

x₁=k + 1, y₁ = 2k;

x₂ = 3k, y₂ = 2k + 3 and

x₃ = 5k −  1, y₃=5k

∴ `1/2|(k+1){(2k+3)−(5k)}+(3k){(5k)−(2k)}+(5k−1){(2k)−(2k+3)}|=0`

⇒ (k+1){−3k+3}+(3k){3k}+(5k−1){−3}=0

⇒ −3k² + 3k − 3k + 3 + 9k² − 15k + 3=0

⇒ 6k² − 15k + 6 = 0

⇒ 2k² − 5k + 2 = 0

⇒ 2k² − 4k − k + 2 = 0

⇒ 2k(k − 2) −1(k − 2) = 0

⇒(k − 2)(2k − 1) = 0

`=> k = 1/2, 2`

So the value of k are `1/2` and 2