Ex 7.1, 9 - If Q(0, 1) is equidistant from P(5, -3) R(x, 6) - Equidistant points

  1. Chapter 7 Class 10 Coordinate Geometry
  2. Serial order wise
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Ex 7.1 , 9 If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR. Since Q is equidistant from P & R QP = QR Finding QP x1 = 0 , y1 = 1 x2 = 5 , y2 = βˆ’3 QP= √((π‘₯2 βˆ’π‘₯1)2+(𝑦2 βˆ’π‘¦1)2) = √(( 5 βˆ’0)2+(βˆ’3 βˆ’1)2) = √((5)2+(βˆ’4)2) = √(25+16) = √41 Similarly, Finding QR x1 = 0 , y1 = 1 x2 = x , y2 = 6 QR= √((π‘₯2 βˆ’π‘₯1)2+(𝑦2 βˆ’π‘¦1)2) = √(( π‘₯ βˆ’0)2+(6 βˆ’1)2) = √((π‘₯)2+(5)2) = √(π‘₯2+ 25) Since , QP = QR √41 = √(π‘₯2+ 25) Squaring both sides (√41)2 = (√(π‘₯2+ 25)) 2 41 = x 2 + 25 0 = x 2 + 25 βˆ’ 41 0 = x 2 βˆ’ 16 x 2 βˆ’ 16 = 0 x 2 = 0 + 16 x 2 = 16 x = Β± √16 x = Β± 4 So, x = 4 or x = βˆ’4 Therefore, point R(x, 6) is (4, 6) or (βˆ’4, 6) Now we need to find the distances PR & QR Finding QR QR = √(π‘₯2+ 25) Hence QR = √41 Finding PR x1 = 5 , y1 = βˆ’3 x2 = x , y2 = 6 PR =√((π‘₯ βˆ’5)2+(6 βˆ’(βˆ’3))2) = √((π‘₯ βˆ’5)2+(6+3)2) = √((π‘₯ βˆ’5)2+(9)2) Hence PR = √82 or √162

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