Check sibling questions

Ex 10.3, 12 - Two lines passing through (2, 3) intersects

Ex 10.3, 12 - Chapter 10 Class 11 Straight Lines - Part 2
Ex 10.3, 12 - Chapter 10 Class 11 Straight Lines - Part 3 Ex 10.3, 12 - Chapter 10 Class 11 Straight Lines - Part 4 Ex 10.3, 12 - Chapter 10 Class 11 Straight Lines - Part 5

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Transcript

Ex 9.3, 11 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 Taking (𝒎_𝟐 − 𝟐)/(𝟏 + 𝟐𝒎_𝟐 ) = √𝟑 m2 − 2 = √3(1 + 2m2) m2 − 2 = √3 + 2√3m2 m2 – 2√3m2 = √3 + 2 m2 (1 – 2√3) = 2 + √3 m2 = (2 + √3)/(1 − 2√3) Taking (𝒎_𝟐 − 𝟐)/(𝟏 + 𝟐𝒎_𝟐 ) = − √𝟑 m2 − 2 = − √3(1 + 2m2) m2 − 2 = − √3 − 2√3m2 m2 + 2√3m2 = −√3 + 2 m2 (1 + 2√3) = 2 − √3 m2 = (2 − √3)/(1 + 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.