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Ex 7.2, 9 Find the coordinates of the points which divide the line segment joining A(โ€“ 2, 2) and B(2, 8) into four equal parts. Let the points that divide AB into 4 equal Parts be P1, P2 and P3 We know that AP1 = P1P2 = P2P3 = P3B Assuming AP1 = P1P2 = P2P3 = P3B = k Hence ๐ด๐‘ƒ2/๐‘ƒ2๐ต = (๐ด๐‘ƒ1 + ๐‘ƒ1๐‘ƒ2)/(๐‘ƒ2๐‘ƒ3 +๐‘ƒ3๐ต) = ( ๐‘˜ + ๐‘˜)/(๐‘˜ + ๐‘˜) = ( 2๐‘˜)/2๐‘˜ = 1/1 = 1 : 1 Hence Point P2 divides AB into two equal parts AP2 & P2B Hence the coordinates of P2 are ((๐‘ฅ1 + ๐‘ฅ2)/2 ", " (๐‘ฆ1 +๐‘ฆ2)/2) = ((โˆ’2 + 2)/2 ", " (2 + 8)/2) = (0/2 ", " 10/2) = (0, 5) So, P2 (0, 5) Similarly, ๐ด๐‘ƒ1/๐‘ƒ1๐‘ƒ2 = ( ๐‘˜)/๐‘˜ = 1/1 = 1 : 1 Hence Point P1 divides AP2 into two equal parts Hence the coordinates of P1 are ((๐‘ฅ1 + ๐‘ฅ2)/2 ", " (๐‘ฆ1 +๐‘ฆ2)/2) = ((โˆ’2 +0)/2 "," (2 +5)/2) = ((โˆ’2)/2 ", " 7/2) = ("โˆ’1, " 7/2) So, P1 ("โˆ’1, " ๐Ÿ•/๐Ÿ) Similarly, ๐‘ƒ2๐‘ƒ3/๐‘ƒ3๐ต = ( ๐‘˜)/๐‘˜ = 1/1 = 1 : 1 Hence Point P3 divides P2B into two equal parts Hence the coordinates of P3 are ((๐‘ฅ1 + ๐‘ฅ2)/2 ", " (๐‘ฆ1 +๐‘ฆ2)/2) = ((0 + 2)/2 ", " (5 + 8)/2) = (2/2 ", " 13/2) = ("1, " 13/2) So, P3 ("1, " ๐Ÿ๐Ÿ‘/๐Ÿ)

  1. Chapter 7 Class 10 Coordinate Geometry
  2. Serial order wise

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo