# Ex 12.3, 4

Last updated at March 9, 2017 by Teachoo

Last updated at March 9, 2017 by Teachoo

Transcript

Ex 12.3, 4 Using section formula, show that the points A (2, 3, 4), B ( 1, 2, 1) and C 0, 1 3 ,2 are collinear. Given Points A (2, 3, 4) , B ( 1, 2, 1) & C 0, 1 3 ,2 Point A, B & C are collinear if , point C divides AB in some ratio externally or internally. We know that Co-ordinate of point P(x, y ,z) that divides line segment joining (x1, y1, z1) & (x2, y2, z2) in ration m : n is (x, y ,z) = mx2 + nx1 m + n , my2+ my1 m + n Here, let point C(14, 0, 2) divide A( 4, 6, 10) , B(2, 4, 6) in the ratio k : 1 Here, m = k , n = 1 x1 = 2, y1 = 3, z1 = 4 x2 = 1, y2 = 2, z2 = 1 Putting values 0, 1 3 , 2 = k 1 + 1(2) k + 1 , k 2 + 1(3) k + 1 , k 1 + 1(4) k + 1 0, 1 3 , 2 = k + 2 k + 1 , 2k + 3 k + 1 , k + 4 k + 1 Comparing x coordinate 0 = k + 2 k + 1 0 (k + 1) = k + 2 0 = k + 2 k + 2 = 0 k = 2 k = 2 So, k : 1 = 2 : 1 Hence point C divide line segment AB in the ratio 2 : 1 Hence points A, B & C are collinear

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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.