Question

Find the values of k if the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.

Answer 1

We know that, if three points are collinear, then the area of triangle formed by these points is zero.

Since, the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.

Then, area of ΔABC = 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, x₂ = 3k, x₃ = 5k – 1 and y₁ = 2k₁, y₂ = 2k + 3, y₃ = 5k

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

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

⇒ `1/2[-3k^2 + 3k - 3k + 3 + 9k^2 - 15k + 3]` = 0

⇒ `1/2(6k^2 - 15k + 6)` = 0  ...[Multiply by 2]

⇒ 6k² – 15k + 6 = 0   ...[By factorisation method]

⇒ 2k² – 5k + 2 = 0   ...[Divide by 3]

⇒ 2k² – 4k – k + 2 = 0

⇒ 2k(k – 2) – 1(k – 2) = 0

⇒ (k – 2)(2k – 1) = 0

If k – 2 = 0, then k = 2

If 2k – 1 = 0, then k = `1/2`

∴ k = `2, 1/2`

Hence, the required values of k are 2 and `1/2`.