Q22 of 52 Page 1

Find the value of k, for which the points A(6, –1), B(k, –6) and C(0, –7) are collinear.

Given,


A (6, -1),


B (k -6)


C (0, -7) Points are collinear


∵ Area of triangle = 0


Area of the triangle = 1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)] = 0


Therefore,


6[-6-(-7)]+k[-7-(-1)+0[-1-(-6)] = 0


6 – 6k = 0


- 6k = - 6


k = 1


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