Q32 of 40 Page 1

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

Three points are collinear if the area of triangle formed by them is zero.

Area of a triangle


0 = (k + 1)(2k + 3 – 5k) + 3k(5k – 2k) + (5k – 1)(2k – 2k – 3)


⇒ 0 = (k + 1)(3 – 3k) + 3k(3k) + (5k – 1)(-3)


⇒ 0 = -3k2+ 3 + 9k2 – 15k + 3


⇒ 6k2 – 15k + 6 = 0


⇒ 6k2 – 12k – 3k + 6 = 0


⇒ 6k(k – 2) – 3(k – 2) = 0


⇒ (6k – 3)(k – 2) = 0


⇒ k = 2 or k = 1/2


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