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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 = −3 QP = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 5 −0)2+(−3 −1)2) = √((5)2+(−4)2) = √(25+16) = √41 Similarly, Finding QR x1 = 0, y1 = 1 x2 = x, y2 = 6 QR = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 𝑥 −0)2+(6 −1)2) = √((𝑥)2+(5)2) = √(𝑥2+ 25) Since, QP = QR √41 = √(𝑥2+ 25) Squaring both sides (√41)2 = (√(𝑥2+ 25)) 2 41 = x 2 + 25 0 = x 2 + 25 − 41 0 = x 2 − 16 x 2 − 16 = 0 x 2 = 0 + 16 x 2 = 16 x = ± √16 x = ± 4 So, x = 4 or x = −4 Therefore, point R(x, 6) is (4, 6) or (−4, 6) Now we need to find the distances PR & QR Finding QR QR = √(𝑥2+ 25) Hence, QR = √𝟒𝟏 Taking x = 4 QR = √(𝑥2+ 25) = √(42+ 25) = √(16+ 25) = √𝟒𝟏 Taking x = −4 QR = √(𝑥2+ 25) = √((−4)2+ 25) = √(16+ 25) = √𝟒𝟏 Finding PR x1 = 5, y1 = −3 x2 = x, y2 = 6 PR = √((𝑥 −5)2+(6 −(−3))2) = √((𝑥 −5)2+(6+3)2) = √((𝑥 −5)2+(9)2) Hence, PR = √𝟖𝟐 or 𝟗√𝟐 Taking x = 4 PR = √((𝑥−5)^2+9^2 ) = √((4−5)2+81) = √((−1)2+81) = √(1+81) = √82 Taking x = –4 PR = √((𝑥−5)^2+9^2 ) = √((−4−5)2+81) = √((−9)2+81) = √(81+81) = 9√2

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.