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2. 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)

Harshit Singh , 4 Years ago
Grade 12th pass
anser 1 Answers
Pawan Prajapati
(i) For collinear points, area of triangle formed by them is always zero. Let points (7, -2) (5, 1), and (3, k) are vertices of a triangle. Area of triangle = 1/2 [7 { 1- k} + 5(k-(-2)) + 3{(-2) – 1}] = 0 7 – 7k + 5k +10 -9 = 0 -2k + 8 = 0 k = 4 (ii) For collinear points, area of triangle formed by them is zero. Therefore, for points (8, 1), (k, – 4), and (2, – 5), area = 0 1/2 [8 { -4- (-5)} + k{(-5)-(1)} + 2{1 -(-4)}] = 0 8 – 6k + 10 = 0 6k = 18 k = 3
Last Activity: 4 Years ago
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