Ex 7.1, 7 - Find point on x−axis which is equidistant from - Ex 7.1

  1. Chapter 7 Class 10 Coordinate Geometry
  2. Serial order wise
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Ex 7.1 , 7 Find the point on the x axis which is equidistant from (2, 5) and ( 2, 9). Let the given points be P(2, 5) , Q( 2, 9) And the point required be R(a, 0) Note: Since the point is on the x axis, y = 0. And assuming x = a Hence the point is (a, 0) As per question, point R is equidistant from P & Q Hence, PR = QR Finding PR x1 = 2 , y1 = 5 x2 = a , y2 = 0 PR = (( 2 1)2+( 2 1)2) = (( 2)2+(0 ( 5))2) = (( 2)2+(5)2) = ( 2+22 2(2)( )+(5)2) = ( 2+4 4 +25) = ( 2 4 +29) Finding QR x1 = 2 , y1 = 9 x2 = a , y2 = 0 QR = (( 2 1)2+( 2 1)2) = (( ( 2))2+(0 (9))2) = (( +2)2+( 9)2) = ( 2+22+2(2)( )+(9)2) = ( 2+4+4 +81) = ( 2+4 +85) From the question PR = QR ( 2 4 +29) = ( 2+4 +85) Squaring both sides ( ( 2 4 +29) )2 = ( ( 2+4 +85))2 a2 4a + 29 = a2 + 4a + 85 a2 4a a2 4a = 85 29 8a = 56 a = 56/( 8) a = 7 Hence the required point is R(a, 0) i.e. ( 7, 0)

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