Ex 7.2, 9 - Find coordinates of points which divide A(-2, 2) - Ex 7.2

part 2 - Ex 7.2, 9 - Ex 7.2 - Serial order wise - Chapter 7 Class 10 Coordinate Geometry
part 3 - Ex 7.2, 9 - Ex 7.2 - Serial order wise - Chapter 7 Class 10 Coordinate Geometry part 4 - Ex 7.2, 9 - Ex 7.2 - Serial order wise - Chapter 7 Class 10 Coordinate Geometry part 5 - Ex 7.2, 9 - Ex 7.2 - Serial order wise - Chapter 7 Class 10 Coordinate Geometry part 6 - Ex 7.2, 9 - Ex 7.2 - Serial order wise - Chapter 7 Class 10 Coordinate Geometry

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Ex 7.2, 9 Find the coordinates of the points which divide the line segment joining A(– 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 š“š‘ƒ2/š‘ƒ2šµ = (š“š‘ƒ1 + š‘ƒ1š‘ƒ2)/(š‘ƒ2š‘ƒ3 +š‘ƒ3šµ) = ( š‘˜ + š‘˜)/(š‘˜ + š‘˜) = ( 2š‘˜)/2š‘˜ = 1/1 = 1 : 1 Hence Point P2 divides AB into two equal parts AP2 & P2B Hence the coordinates of P2 are ((š‘„1 + š‘„2)/2 ", " (š‘¦1 +š‘¦2)/2) = ((āˆ’2 + 2)/2 ", " (2 + 8)/2) = (0/2 ", " 10/2) = (0, 5) So, P2 (0, 5) Similarly, š“š‘ƒ1/š‘ƒ1š‘ƒ2 = ( š‘˜)/š‘˜ = 1/1 = 1 : 1 Hence Point P1 divides AP2 into two equal parts Hence the coordinates of P1 are ((š‘„1 + š‘„2)/2 ", " (š‘¦1 +š‘¦2)/2) = ((āˆ’2 +0)/2 "," (2 +5)/2) = ((āˆ’2)/2 ", " 7/2) = ("āˆ’1, " 7/2) So, P1 ("āˆ’1, " šŸ•/šŸ) Similarly, š‘ƒ2š‘ƒ3/š‘ƒ3šµ = ( š‘˜)/š‘˜ = 1/1 = 1 : 1 Hence Point P3 divides P2B into two equal parts Hence the coordinates of P3 are ((š‘„1 + š‘„2)/2 ", " (š‘¦1 +š‘¦2)/2) = ((0 + 2)/2 ", " (5 + 8)/2) = (2/2 ", " 13/2) = ("1, " 13/2) So, P3 ("1, " šŸšŸ‘/šŸ)

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