# Ex 10.3, 12

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 10.3, 12 Two lines passing through the point (2, 3) intersects each other at an angle of 60°. If slope of one line is 2, find equation of the other line. We know that Angle between 2 lines be tan θ =|(𝑚_2 − 𝑚_1)/(1 + 𝑚_2 𝑚_1 )| Here m1 = Slope of one line = 2 θ = 60° (given) We need to find m2 Putting the values tan 60° = |(𝑚_2 − 2)/(1 + 2 × 𝑚_2 )| √3 = |(𝑚_2 − 2)/(1 + 2𝑚_2 )| |(𝑚_2 − 2)/(1 + 2𝑚_2 )|= √3 (𝑚_2 − 2)/(1 + 2𝑚_2 ) = ± √3 So, (𝑚_2 − 2)/(1 + 2𝑚_2 ) = √3 and (𝑚_2 − 2)/(1 + 2𝑚_2 ) = – √3 We know that equation of a line passing through (x1, y1) & having slope m is (y − y1) = m(x − x1) Equation of a line passing through (2, 3) & having slope (𝟐 + √𝟑)/(𝟏 − 𝟐√𝟑) is (y − 3) = ( (2 + √3))/(1 − 2√3) (x − 2) (1 − 2√3)(y − 3) = (2 + √3)(x − 2) 1(y − 3) − 2√3(y − 3) = 2(x – 2) + √3(x − 2) y – 3 − 2√3y + 6√3 = 2x – 4 + √3x − 2√3 y − 2√3y − 4x − √3x = – 6√3 – 2√3 – 4 + 3 y (1 − 2√3) − x(√3 + 2) = – 1 – 8√3 1 + 8√3 = x(√3 + 2) + y (2√3 – 1) (√3 + 2)x + (2√3 – 1)y = 1 + 8√3 Equation of a line passing through (2, 3) & having slope (𝟐 − √𝟑)/(𝟏 + 𝟐√𝟑) is (y − 3) = (2 − √3)/(2√3 + 1)(x − 2) (2√3 + 1) (y − 3) = (2 − √3) (x − 2) 2√3 (y − 3) + 1 (y − 3) = 2(x − 2) − √3(x − 2) 2√3 (y − 3) + 1 (y − 3) = 2(x − 2) − √3(x − 2) 2√3y − 6√3 + y − 3 = 2x − 4 − √3x + 2√3 2√3y + y − 6√3 − 3 = 2x − √3x − 4 + 2√3 (2√3 + 1)y − 6√3 − 3 = (2 − √3)x − 4 + 2√3 (2√3 + 1)y = − (√3 − 2)x − 4 + 2√3 + 6√3 + 3 (2√3 + 1)y + (√3 − 2)x = 2√3 + 6√3 − 4 + 3 (√3 − 2)x + (2√3 + 1)y = 8√3 − 1 Hence the equation of lines is (√3 + 2)x + (2√3 – 1)y = 1 + 8√3 or (√3 − 2)x + (2√3 + 1)y = 8√3 − 1

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.