Q31 of 46 Page 1

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

As A, B and C are collinear


Then,


ar(△ABC) = 0


Using formula for a triangle if three points (x1, y1), (x2, y2) and (x3, y3) are given.



We have,



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


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


6k2 - 15k + 6 = 0


2k2 - 5k + 2 = 0


2k2 - 4k - k + 2 = 0


2k(k - 2) - 1(k - 2) = 0


(2k - 1)(k - 2) = 0



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