Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Equidistant points
Last updated at May 29, 2023 by Teachoo
Ex 7.1, 9 If Q(0, 1) is equidistant from P(5, β3) 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 = β3 QP = β((π₯2 βπ₯1)2+(π¦2 βπ¦1)2) = β(( 5 β0)2+(β3 β1)2) = β((5)2+(β4)2) = β(25+16) = β41 Similarly, Finding QR x1 = 0, y1 = 1 x2 = x, y2 = 6 QR = β((π₯2 βπ₯1)2+(π¦2 βπ¦1)2) = β(( π₯ β0)2+(6 β1)2) = β((π₯)2+(5)2) = β(π₯2+ 25) Since, QP = QR β41 = β(π₯2+ 25) Squaring both sides (β41)2 = (β(π₯2+ 25)) 2 41 = x 2 + 25 0 = x 2 + 25 β 41 0 = x 2 β 16 x 2 β 16 = 0 x 2 = 0 + 16 x 2 = 16 x = Β± β16 x = Β± 4 So, x = 4 or x = β4 Therefore, point R(x, 6) is (4, 6) or (β4, 6) Now we need to find the distances PR & QR Finding QR QR = β(π₯2+ 25) Hence, QR = βππ Taking x = 4 QR = β(π₯2+ 25) = β(42+ 25) = β(16+ 25) = βππ Taking x = β4 QR = β(π₯2+ 25) = β((β4)2+ 25) = β(16+ 25) = βππ Finding PR x1 = 5, y1 = β3 x2 = x, y2 = 6 PR = β((π₯ β5)2+(6 β(β3))2) = β((π₯ β5)2+(6+3)2) = β((π₯ β5)2+(9)2) Hence, PR = βππ or πβπ Taking x = 4 PR = β((π₯β5)^2+9^2 ) = β((4β5)2+81) = β((β1)2+81) = β(1+81) = β82 Taking x = β4 PR = β((π₯β5)^2+9^2 ) = β((β4β5)2+81) = β((β9)2+81) = β(81+81) = 9β2