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Ex 7.3 , 2 In each of the following find the value of ‘k’, for which the points are collinear. (7, –2), (5, 1), (3, k) Let the given points be A (7, −2), B (5, 1), C (3, k) If the above points are collinear, they will lie on the same line, i.e. the will not form triangle or We can say that Area of ∆ABC = 0 1/2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0 Here x1 = 7, y1 = −2 x2 = 5, y2 = 1 x3 = 3, y3 = k Putting values 1/2 [7(1 − k) + 5(k – (−2)) + 3(−2 − 1)] = 0 7(1 − k) + 5(k +2) + 3(−3) = 0 × 2 7 – 7k + 5k + 10 – 9 = 0 – 7k + 5k = − 10 + 9 – 7 −2k = −8 −2k = −8 k = (−8)/(−2) k = 4

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.