# 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 0,13,2 are collinear. Given Points A (2, –3, 4) , B (–1, 2, 1) & C 0, 13 ,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 + nx1m + n,my2+ my1m + 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, 13, 2 =k−1 + 1(2)k + 1,k2 + 1(3)k + 1,k1 + 1(4)k + 1 0, 13, 2 =− k + 2k + 1,2k + 3k + 1,k + 4k + 1 Comparing x – coordinate 0 = −k + 2k + 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

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