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

Transcript

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)

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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.