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Misc 19 - If lines y = 3x + 1, 2y = x + 3 are equally inclined

Misc 19 - Chapter 10 Class 11 Straight Lines - Part 2
Misc 19 - Chapter 10 Class 11 Straight Lines - Part 3 Misc 19 - Chapter 10 Class 11 Straight Lines - Part 4 Misc 19 - Chapter 10 Class 11 Straight Lines - Part 5 Misc 19 - Chapter 10 Class 11 Straight Lines - Part 6 Misc 19 - Chapter 10 Class 11 Straight Lines - Part 7 Misc 19 - Chapter 10 Class 11 Straight Lines - Part 8


Transcript

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 ∴ 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 ∴ 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‘š_2 βˆ’ π‘š_1)/(1 + π‘š_2 π‘š_1 )| Finding angle between AB & PQ Slope of AB = m1 = 3 & slope of PQ = m tan ΞΈ = |(π‘š βˆ’ 3)/(1 + 3π‘š)| |(π‘š βˆ’ 3)/(1 + 3π‘š)| = tan ΞΈ Finding angle between CD & PQ Slope of PR = 1/2 & slope of PQ = m tan ΞΈ = |(π‘š βˆ’ 1/2)/(1 + 1/2 π‘š)| tan ΞΈ = |((2π‘š βˆ’ 1)/2)/((2 + π‘š)/2)| = |(2π‘š βˆ’ 1)/(2 + π‘š)| Since angles (ΞΈ) are equal So, tan ΞΈ must be equal From (A) & (B) |(π‘š βˆ’ 3)/(1 + 3π‘š)| = |(2π‘š βˆ’ 1)/(2 + π‘š)| So, ((π‘š βˆ’ 3)/(1 + 3π‘š)) = ((2π‘š βˆ’ 1)/(2 + π‘š)) or ((π‘š βˆ’ 3)/(1 + 3π‘š)) = βˆ’ ((2π‘š βˆ’ 1)/(2 + π‘š)) Solving ((π’Ž βˆ’ πŸ‘)/(𝟏 + πŸ‘π’Ž))= ((πŸπ’Ž βˆ’ 𝟏)/(𝟐 + π’Ž)) (π‘š βˆ’ 3)/(1 + 3π‘š) = (2π‘š βˆ’ 1)/(2 + π‘š) (m βˆ’ 3)(2 + m) = (2m βˆ’ 1) (1 + 3m) m (2 + m) βˆ’ 3(2 + m) = 2m + 6m2 βˆ’ 1 + 3m 5m + m2 βˆ’ 6 = 5m + 6m2 βˆ’ 1 5m βˆ’ 5m + m2 βˆ’ 6m2 βˆ’ 6 + 1 = 0 0 βˆ’ 5m2 βˆ’ 5 = 0 βˆ’ 5m2 = 5 m2 = 5/( βˆ’ 5) m2 = βˆ’ 1 This is not possible as square of a number cannot be negative Solving (π’Ž βˆ’ πŸ‘)/(𝟏 + πŸ‘π’Ž) = βˆ’ ((πŸπ’Ž βˆ’ 𝟏)/(𝟐 + π’Ž)) (π‘š βˆ’ 3)/(1 + 3π‘š) = (1 βˆ’ 2π‘š)/(2 + π‘š) (m βˆ’ 3) (2 + m) = (1 βˆ’ 2m)(1 + 3m) m(2 + m) βˆ’ 3(2 + m) = 1(1 + 3m) βˆ’ 2m(1 + 3m) 2m + m2 βˆ’ 6 βˆ’ 3m = 1 + 3m βˆ’ 2m βˆ’ 6m2 m2 βˆ’ 6 βˆ’ m = βˆ’ 6m2 + m + 1 m2 + 6m2 βˆ’ m βˆ’ m βˆ’ 6 βˆ’ 1 = 0 7m2 βˆ’ 2m βˆ’ 7 = 0 The above equation is of the form ax2 + 3x + c = 0 Where solution is x = ( βˆ’ 𝑏 Β± √(𝑏^2 βˆ’ 4π‘Žπ‘))/2π‘Ž Here a = 7, b = βˆ’ 2,c = –7 & x = m So, m = ( βˆ’(βˆ’2) Β± √(γ€–(βˆ’2)γ€—^2 βˆ’ 4(7)(βˆ’7)))/(2(7)) m = (2 Β± √(4 (1 + (7) Γ— (βˆ’7)))/14 m = (2 Β± 2 √((1 + 49)))/14 m = (2(1 Β± √50 ))/14 m = (2(1Β± √(5 Γ— 5 Γ— 2)))/14 m = 1/7(1 Β± 5√2) m = (1 Β± 5√2)/7 Thus, the required value of m is (𝟏 + πŸ“βˆšπŸ)/πŸ• & (𝟏 βˆ’ πŸ“βˆšπŸ)/πŸ•

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.