# Ex 7.3, 2 - Chapter 7 Class 10 Coordinate Geometry

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

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 Ex 7.3 , 2 In each of the following find the value of k , for which the points are collinear. (ii) (8, 1), (k, 4), (2, 5) Let the given points be A(8, 1) , B(k, 4) , C(2, 5) 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 = 8 , y1 = 1 x2 = k , y2 = 4 x3 = 2 , y3 = 5 Putting values 1/2 [ 8( 4 ( 5)) + k( 5 1) + 2(1 ( 4)) ] = 0 8( 4 + 5) + k( 6) + 2(1 + 4) = 0 2 8(1) + k( 6) + 2(5) = 0 8 6k + 10 = 0 6k = 10 8 6k = 10 8 6k = 18 k = ( 18)/( 6) k = 3

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.