641 students have Teachoo Black. What are you waiting for?

Miscellaneous

Misc 1
Important

Misc 2 Important

Misc 3 Important

Misc 4

Misc 5

Misc 6 Important

Misc 7

Misc 8 Important

Misc 9 Important

Misc 10

Misc 11 Important

Misc 12 Important

Misc 13

Misc 14

Misc 15 Important

Misc 16 Important

Misc 17

Misc 18 Important

Misc 19 Important You are here

Misc 20 Important

Misc 21 Important

Misc 22 Important

Misc 23

Misc 24 Important

Chapter 10 Class 11 Straight Lines (Term 1)

Serial order wise

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