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  1. Chapter 10 Class 11 Straight Lines
  2. Serial order wise

Transcript

Ex 10.1, 11 The slope of a line is double of the slope of another line. If tangent of the angle between them is 1/3 , find the slopes of the lines. Let m1 & m2 be the slopes of two lines We know that angles between two lines are tan ฮธ = |(๐‘š2 โˆ’ ๐‘š1)/(1 + ๐‘š1๐‘š2)| Here tan ฮธ = 1/3 & m2 = 2m1 Putting values tan ฮธ = |(๐‘š2 โˆ’ ๐‘š1)/(1 + ๐‘š1๐‘š2)| 1/3 = |(2๐‘š1 โˆ’ ๐‘š1)/(1 + ๐‘š1(2๐‘š1))| 1/3 = |๐‘š_1/(1 + 2ใ€–๐‘š_1ใ€—^2 )| |๐‘š_1/(1 + 2ใ€–๐‘š_1ใ€—^2 )| = 1/3 So, ๐‘š_1/(1 + 2ใ€–๐‘š_1ใ€—^2 ) = 1/3 or ๐‘š_1/(1 + 2ใ€–๐‘š_1ใ€—^2 ) = ( โˆ’1)/3 Solving ๐’Ž_๐Ÿ/(๐Ÿ + ๐Ÿใ€–๐’Ž_๐Ÿใ€—^๐Ÿ ) = ๐Ÿ/๐Ÿ‘ 3m1 = 1 + 2ใ€–"m1" ใ€—^2 2ใ€–"m1" ใ€—^2 + 1 โ€“ 3m1 = 0 2ใ€–"m1" ใ€—^2 โ€“ 3m1 + 1 = 0 2ใ€–"m1" ใ€—^2 โ€“ 2m1 โ€“ m1 + 1 = 0 2m1(m1 โ€“ 1) โ€“ 1(m1 โ€“ 1) = 0 (2m1 โ€“ 1) (m1 โ€“ 1) = 0 So, m1 = ๐Ÿ/๐Ÿ , m1 = 1 Solving ๐’Ž_๐Ÿ/(๐Ÿ + ๐Ÿใ€–๐’Ž_๐Ÿใ€—^๐Ÿ ) = (โˆ’๐Ÿ)/๐Ÿ‘ 3m1 = โ€“1 โ€“ 2ใ€–"m1" ใ€—^2 2ใ€–"m1" ใ€—^2 + 1 + 3m1 = 0 2ใ€–"m1" ใ€—^2 + 3m1 + 1 = 0 2ใ€–"m1" ใ€—^2 + 2m1 + m1 + 1 = 0 2m1(m1 + 1) + 1(m1 + 1) = 0 (2m1 + 1) (m1 + 1) = 0 So, m1 = (โˆ’๐Ÿ)/๐Ÿ , m1 = โ€“1 When m1 = ( ๐Ÿ)/๐Ÿ m2 = 2m1 m2 = 2(1/2) = 1 When m1 = 1 m2 = 2m1 m2 = 2(1) = 2 When m1 = ( โˆ’๐Ÿ)/๐Ÿ m2 = 2m1 m2 = 2(( โˆ’ 1)/2) = โ€“1 When m1 = โ€“1 m2 = 2m1 m2 = 2(โˆ’1) = โ€“2 Hence slope of lines are ๐Ÿ/๐Ÿ and 1 or 1 and 2 or ( โˆ’๐Ÿ)/๐Ÿ and โˆ’1 or โˆ’1 and โˆ’2

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.