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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) As per question, point R is equidistant from P & Q Hence, PR = QR Note: Since the point is on the x−axis, y = 0 And assuming x = a Hence the point is (a, 0) Finding PR x1 = 2, y1 = −5 x2 = a, y2 = 0 PR = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 𝑎 −2)2+(0−(−5))2) = √((𝑎−2)2+(5)2) = √(𝑎2+22 −2(2)(𝑎)+(5)2) = √(𝑎2+4 −4𝑎+25) = √(𝑎2 −4𝑎+29) Finding QR x1 = −2, y1 = 9 x2 = a, y2 = 0 QR = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) = √(( 𝑎 −(−2))2+(0−(9))2) = √((𝑎+2)2+(−9)2) = √(𝑎2+22+2(2)(𝑎)+(9)2) = √(𝑎2+4+4𝑎+81) = √(𝑎2+4𝑎+85) From the question PR = QR √(𝑎2 −4𝑎+29) = √(𝑎2+4𝑎+85) Squaring both sides (√(𝑎2 −4𝑎+29) )2 = (√(𝑎2+4𝑎+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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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.