Question

Please answer it in a page ....
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Answer 1

Note: I don't have pen and paper with me at present..Sorry!

Given points are A(k + 1, 2k), B(3k, 2k + 3), C(5k - 1, 5k).

Here,  x₁ = k + 1, x₂ = 3k, x₃ = 5k - 1.

          y₁ = 2k, y₂ = 2k + 3, y₃ = 5k.

Given that the three points are collinear.

Area of ΔABC = 0

⇒ x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) = 0

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

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

⇒ -3k^2 + 3k - 3k + 3 + 9k^2 - 15k + 3 = 0

⇒ 6k^2 - 15k + 6 = 0

⇒ 3(2k^2 - 5k + 2) = 0

⇒ 2k^2 - 5k + 2 = 0

⇒ 2k^2 - 4k - k + 2 = 0

⇒ 2k(k - 2) - (k - 2) = 0

⇒ (k - 2)(2k - 1) = 0

⇒ k = 2, 1/2

Therefore, the value of k = 2, 1/2.

Hope it helps!