Answer :

Given:

A (say) = (–5, 1)


B (say) = (1, k)


C (say) = (4, –2)


For A, B and C to be collinear they would lie in same line and therefore cannot make triangle.



i.e.,


Area of ∆ABC should be 0.


We know that,


Area of ∆ABC = 1/2[x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]


Area of ∆ABC = 1/2[(-5)(k – (-2)) + 1((-2) – 1) + 4(1 – k)] = 0


1/2[(-5)(k + 2) + 1(-3) + 4(1 – k)] = 0


1/2[-5k – 10 – 3 + 4 – 4k] = 0


-5k 10 3 + 4 4k = 0


-9k – 9 = 0


-9(k + 1) = 0


k + 1 = 0


k = -1


Hence, k = -1


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