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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

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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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.