In each of the following find the value of ‘k’, for which the points are collinear. (i) (7, –2), (5, 1), (3, k) (ii) (8, 1), (k, – 4), (2, –5)

(i) A(7,-2), B(5, 1), C(3,k)
Area of triangle formed by three collinear points is zero.
Therefore, the area of triangle ABC = 0
⇒ 1/2[7(1-k) +5{k- (-2)) +3(-2-1))] =0
⇒ 7-7k+ 5k + 10 -9 = 0
⇒ -2k = -8
⇒ k = 4
(ii) P(8, 1), Q(k,-4), R(2,-5)
Area of triangle formed by three collinear points is zero.
Therefore, the area of triangle PQR = 0
⇒ 1/2[8{-4-(-5)+k(-5-1) + 2{1 -(-4)}] = 0
⇒ 8-6k+10 = 0
⇒ -6k = -18
⇒ k = 3

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