Q28 of 47 Page 1

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

Let the three points be A(k + 1, 2k) B(3k, 2k + 3) and C(5k - 1, 5k)


Three points A, B and C are collinear if and only if


Area(△ABC) = 0


As we know area of triangle formed by three points (x1, y1) , (x2,y2) and (x3, y3)





⇒ [-3k2 - 3k - 3k - 3 + 9k2 - 15k + 3] = 0


⇒ 6k2 - 21k = 0


⇒ 6k2 = 21k


⇒ 2k = 7



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