# Example 14 - Chapter 7 Class 10 Coordinate Geometry

Last updated at Feb. 25, 2017 by Teachoo

Last updated at Feb. 25, 2017 by Teachoo

Transcript

Example 14 Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are collinear. If the above points are collinear, they will lie on the same line, i.e. they will not form triangle Therefore, Area of ∆ABC = 0 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] = 0 Here x1 = 2 , y1 = 3 x2 = 4 , y2 = k x3 = 6 , y3 = −3 Putting values 1/2 [ 2(k – (−3)) + 4(−3 − 3) + 6(3 – k) ] = 0 2(k + 3) + 4(−6) + 6(3 − k) = 0 × 2 2k + 6 – 24 + 18 – 6k = 0 2k – 6k = − 6 + 24 −18 −4k = 0 ⇒ k = 0

Chapter 7 Class 10 Coordinate Geometry

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.