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Ex 10.3
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Ex 10.3
Last updated at Feb. 3, 2020 by Teachoo
Ex 10.3, 12 Two lines passing through the point (2, 3) intersects each other at an angle of 60Β°. If slope of one line is 2, find equation of the other line. We know that Angle between 2 lines be tan ΞΈ =|(π_2 β π_1)/(1 + π_2 π_1 )| Here m1 = Slope of one line = 2 ΞΈ = 60Β° (given) We need to find m2 Putting the values tan 60Β° = |(π_2 β 2)/(1 + 2 Γ π_2 )| β3 = |(π_2 β 2)/(1 + 2π_2 )| |(π_2 β 2)/(1 + 2π_2 )|= β3 (π_2 β 2)/(1 + 2π_2 ) = Β± β3 So, (π_2 β 2)/(1 + 2π_2 ) = β3 and (π_2 β 2)/(1 + 2π_2 ) = β β3 Taking (π_π β π)/(π + ππ_π ) = βπ m2 β 2 = β3(1 + 2m2) m2 β 2 = β3 + 2β3m2 m2 β 2β3m2 = β3 + 2 m2 (1 β 2β3) = 2 + β3 m2 = (2 + β3)/(1 β 2β3) Taking (π_π β π)/(π + ππ_π ) = β βπ m2 β 2 = β β3(1 + 2m2) m2 β 2 = β β3 β 2β3m2 m2 + 2β3m2 = ββ3 + 2 m2 (1 + 2β3) = 2 β β3 m2 = (2 β β3)/(1 + 2β3) We know that equation of a line passing through (x1, y1) & having slope m is (y β y1) = m(x β x1) Equation of a line passing through (2, 3) & having slope (π + βπ)/(π β πβπ) is (y β 3) = ( (2 + β3))/(1 β 2β3) (x β 2) (1 β 2β3)(y β 3) = (2 + β3)(x β 2) 1(y β 3) β 2β3(y β 3) = 2(x β 2) + β3(x β 2) y β 3 β 2β3y + 6β3 = 2x β 4 + β3x β 2β3 y β 2β3y β 4x β β3x = β 6β3 β 2β3 β 4 + 3 y (1 β 2β3) β x(β3 + 2) = β 1 β 8β3 1 + 8β3 = x(β3 + 2) + y (2β3 β 1) (βπ + 2)x + (2βπ β 1)y = 1 + 8βπ Equation of a line passing through (2, 3) & having slope (π β βπ)/(π + πβπ) is (y β 3) = (2 β β3)/(2β3 + 1)(x β 2) (2β3 + 1) (y β 3) = (2 β β3) (x β 2) 2β3 (y β 3) + 1 (y β 3) = 2(x β 2) β β3(x β 2) 2β3 (y β 3) + 1 (y β 3) = 2(x β 2) β β3(x β 2) 2β3y β 6β3 + y β 3 = 2x β 4 β β3x + 2β3 2β3y + y β 6β3 β 3 = 2x β β3x β 4 + 2β3 (2β3 + 1)y β 6β3 β 3 = (2 β β3)x β 4 + 2β3 (2β3 + 1)y = β (β3 β 2)x β 4 + 2β3 + 6β3 + 3 (2β3 + 1)y + (β3β2)x = 2β3 + 6β3 β 4 + 3 (βπβπ)x + (2βπ + 1)y = 8βπ β 1 Hence the equation of lines is (β3 + 2)x + (2β3 β 1)y = 1 + 8β3 or (β3 β 2)x + (2β3 + 1)y = 8β3 β 1