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  1. Chapter 7 Class 10 Coordinate Geometry (Term 1)
  2. Serial order wise

Transcript

Ex 7.1, 9 If Q(0, 1) is equidistant from P(5, โ€“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 = โˆ’3 QP = โˆš((๐‘ฅ2 โˆ’๐‘ฅ1)2+(๐‘ฆ2 โˆ’๐‘ฆ1)2) = โˆš(( 5 โˆ’0)2+(โˆ’3 โˆ’1)2) = โˆš((5)2+(โˆ’4)2) = โˆš(25+16) = โˆš41 Similarly, Finding QR x1 = 0, y1 = 1 x2 = x, y2 = 6 QR = โˆš((๐‘ฅ2 โˆ’๐‘ฅ1)2+(๐‘ฆ2 โˆ’๐‘ฆ1)2) = โˆš(( ๐‘ฅ โˆ’0)2+(6 โˆ’1)2) = โˆš((๐‘ฅ)2+(5)2) = โˆš(๐‘ฅ2+ 25) Since, QP = QR โˆš41 = โˆš(๐‘ฅ2+ 25) Squaring both sides (โˆš41)2 = (โˆš(๐‘ฅ2+ 25)) 2 41 = x 2 + 25 0 = x 2 + 25 โˆ’ 41 0 = x 2 โˆ’ 16 x 2 โˆ’ 16 = 0 x 2 = 0 + 16 x 2 = 16 x = ยฑ โˆš16 x = ยฑ 4 So, x = 4 or x = โˆ’4 Therefore, point R(x, 6) is (4, 6) or (โˆ’4, 6) Now we need to find the distances PR & QR Finding QR QR = โˆš(๐‘ฅ2+ 25) Hence, QR = โˆš๐Ÿ’๐Ÿ Taking x = 4 QR = โˆš(๐‘ฅ2+ 25) = โˆš(42+ 25) = โˆš(16+ 25) = โˆš๐Ÿ’๐Ÿ Taking x = โˆ’4 QR = โˆš(๐‘ฅ2+ 25) = โˆš((โˆ’4)2+ 25) = โˆš(16+ 25) = โˆš๐Ÿ’๐Ÿ Finding PR x1 = 5, y1 = โˆ’3 x2 = x, y2 = 6 PR = โˆš((๐‘ฅ โˆ’5)2+(6 โˆ’(โˆ’3))2) = โˆš((๐‘ฅ โˆ’5)2+(6+3)2) = โˆš((๐‘ฅ โˆ’5)2+(9)2) Hence, PR = โˆš๐Ÿ–๐Ÿ or ๐Ÿ—โˆš๐Ÿ Taking x = 4 PR = โˆš((๐‘ฅโˆ’5)^2+9^2 ) = โˆš((4โˆ’5)2+81) = โˆš((โˆ’1)2+81) = โˆš(1+81) = โˆš82 Taking x = โ€“4 PR = โˆš((๐‘ฅโˆ’5)^2+9^2 ) = โˆš((โˆ’4โˆ’5)2+81) = โˆš((โˆ’9)2+81) = โˆš(81+81) = 9โˆš2

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.