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Last updated at Feb. 4, 2020 by Teachoo
Misc 19 If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m. Let line AB be y = 3x + 1 ,line CD be 2y = x + 3 & line PQ be y = mx + 4 Lines AB & CD are equally inclined to the line PQ First we find slopes of lines Slope of line AB y = 3x + 1 The above equation is of the form y = m1x + c1 where slope of line AB = m1 β΄ Slope of line AB = m1 = 3 Slope of line CD 2y = x + 3 y = (π₯ + 3)/2 y = (1/2)x + 3/2 The above equation of the form y = m2x + c2 where slope of line CD = m2 β΄ Slope of CD = m2 = 1/2 Slope of line PQ y = mx + 4 The above equation of the form y = mx + c where m is slope Thus, Slope of line PQ = m Now, given that lines AB & CD are equally inclined to the line PQ i.e. line AB & line CD make equal angle with line PQ Angle between AB & PQ = angle between CD & PQ We know that angle between two lines whose slope are m1 & m2 is tan ΞΈ = |(π_2 β π_1)/(1 + π_2 π_1 )| Finding angle between AB & PQ Slope of AB = m1 = 3 & slope of PQ = m tan ΞΈ = |(π β 3)/(1 + 3π)| |(π β 3)/(1 + 3π)| = tan ΞΈ Finding angle between CD & PQ Slope of PR = 1/2 & slope of PQ = m tan ΞΈ = |(π β 1/2)/(1 + 1/2 π)| tan ΞΈ = |((2π β 1)/2)/((2 + π)/2)| = |(2π β 1)/(2 + π)| Since angles (ΞΈ) are equal So, tan ΞΈ must be equal From (A) & (B) |(π β 3)/(1 + 3π)| = |(2π β 1)/(2 + π)| So, ((π β 3)/(1 + 3π)) = ((2π β 1)/(2 + π)) or ((π β 3)/(1 + 3π)) = β ((2π β 1)/(2 + π)) Solving ((π β π)/(π + ππ))= ((ππ β π)/(π + π)) (π β 3)/(1 + 3π) = (2π β 1)/(2 + π) (m β 3)(2 + m) = (2m β 1) (1 + 3m) m (2 + m) β 3(2 + m) = 2m + 6m2 β 1 + 3m 5m + m2 β 6 = 5m + 6m2 β 1 5m β 5m + m2 β 6m2 β 6 + 1 = 0 0 β 5m2 β 5 = 0 β 5m2 = 5 m2 = 5/( β 5) m2 = β 1 This is not possible as square of a number cannot be negative Solving (π β π)/(π + ππ) = β ((ππ β π)/(π + π)) (π β 3)/(1 + 3π) = (1 β 2π)/(2 + π) (m β 3) (2 + m) = (1 β 2m)(1 + 3m) m(2 + m) β 3(2 + m) = 1(1 + 3m) β 2m(1 + 3m) 2m + m2 β 6 β 3m = 1 + 3m β 2m β 6m2 m2 β 6 β m = β 6m2 + m + 1 m2 + 6m2 β m β m β 6 β 1 = 0 7m2 β 2m β 7 = 0 The above equation is of the form ax2 + 3x + c = 0 Where solution is x = ( β π Β± β(π^2 β 4ππ))/2π Here a = 7, b = β 2,c = β7 & x = m So, m = ( β(β2) Β± β(γ(β2)γ^2 β 4(7)(β7)))/(2(7)) m = (2 Β± β(4 (1 + (7) Γ (β7)))/14 m = (2 Β± 2 β((1 + 49)))/14 m = (2(1 Β± β50 ))/14 m = (2(1Β± β(5 Γ 5 Γ 2)))/14 m = 1/7(1 Β± 5β2) m = (1 Β± 5β2)/7 Thus, the required value of m is (π + πβπ)/π & (π β πβπ)/π