Ex 9.1, 10 - Slope of a line is double of slope of another - Ex 9.1

part 2 - Ex 9.1, 10 - Ex 9.1 - Serial order wise - Chapter 9 Class 11 Straight Lines
part 3 - Ex 9.1, 10 - Ex 9.1 - Serial order wise - Chapter 9 Class 11 Straight Lines

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Ex 9.1, 10 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 Īø = |(š‘š2 āˆ’ š‘š1)/(1 + š‘š1š‘š2)| Here tan Īø = 1/3 & m2 = 2m1 Putting values tan Īø = |(š‘š2 āˆ’ š‘š1)/(1 + š‘š1š‘š2)| 1/3 = |(2š‘š1 āˆ’ š‘š1)/(1 + š‘š1(2š‘š1))| 1/3 = |š‘š_1/(1 + 2ć€–š‘š_1怗^2 )| |š‘š_1/(1 + 2ć€–š‘š_1怗^2 )| = 1/3 So, š‘š_1/(1 + 2ć€–š‘š_1怗^2 ) = 1/3 or š‘š_1/(1 + 2ć€–š‘š_1怗^2 ) = ( āˆ’1)/3 Solving š’Ž_šŸ/(šŸ + šŸć€–š’Ž_šŸć€—^šŸ ) = šŸ/šŸ‘ 3m1 = 1 + 2怖"m1" 怗^2 2怖"m1" 怗^2 + 1 – 3m1 = 0 2怖"m1" 怗^2 – 3m1 + 1 = 0 2怖"m1" 怗^2 – 2m1 – m1 + 1 = 0 2m1(m1 – 1) – 1(m1 – 1) = 0 (2m1 – 1) (m1 – 1) = 0 So, m1 = šŸ/šŸ , m1 = 1 Solving š’Ž_šŸ/(šŸ + šŸć€–š’Ž_šŸć€—^šŸ ) = (āˆ’šŸ)/šŸ‘ 3m1 = –1 – 2怖"m1" 怗^2 2怖"m1" 怗^2 + 1 + 3m1 = 0 2怖"m1" 怗^2 + 3m1 + 1 = 0 2怖"m1" 怗^2 + 2m1 + m1 + 1 = 0 2m1(m1 + 1) + 1(m1 + 1) = 0 (2m1 + 1) (m1 + 1) = 0 So, m1 = (āˆ’šŸ)/šŸ , m1 = –1 When m1 = ( šŸ)/šŸ m2 = 2m1 m2 = 2(1/2) = 1 When m1 = 1 m2 = 2m1 m2 = 2(1) = 2 When m1 = ( āˆ’šŸ)/šŸ m2 = 2m1 m2 = 2(( āˆ’ 1)/2) = –1 When m1 = –1 m2 = 2m1 m2 = 2(āˆ’1) = –2 Hence slope of lines are šŸ/šŸ and 1 or 1 and 2 or ( āˆ’šŸ)/šŸ and āˆ’1 or āˆ’1 and āˆ’2

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo