Example 8 - Using section formula, prove that three points - Examples

  1. Chapter 12 Class 11 Introduction to Three Dimensional Geometry
  2. Serial order wise

Transcript

Example 8 Using section formula, prove that the three points (– 4, 6, 10), (2, 4, 6) and (14, 0, –2) are collinear. Let points be A (– 4, 6, 10) , B (2, 4, 6) , C (14, 0, – 2) Point A, B, & C are collinear if point C divides AB in some ratio externally & 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, x1 = – 4, y1 = 6, z1 = 10 x2 = 2, y2 = 4, z2 = 6 & m = k , n = 1 Putting values (14, 0, – 2) = ﷐﷐𝑘﷐2﷯+1(−4)﷮𝑘+1﷯,﷐𝑘﷐4﷯+1 (6)﷮𝑘+1﷯,﷐𝑘 ﷐6﷯+1 (10)﷮𝑘+1﷯﷯ (14, 0, – 2) = ﷐﷐2𝑘 −4 ﷮ 𝑘+1﷯,﷐4𝑘+ 6﷮k +1﷯,﷐6𝑘+10﷮ k + 1﷯﷯ Comparing x – coordinate 14 = ﷐2𝑘 − 4﷮𝑘+1﷯ (k + 1) (14) = 2k – 4 14 k + 14 = 2k – 4 14 k – 2k = – 4 – 14 12 k = – 18 k = ﷐−18﷮12﷯ k = ﷐−3﷮2﷯ Since k is negative Point C divide line segment AB externally in the ratio 3 : 2 Thus, A, B & C are collinear

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