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


Transcript

Ex 10.3, 12 Two lines passing through the point (2, 3) intersects each other at an angle of 60Β°. If slope of one line is 2, find equation of the other line. We know that Angle between 2 lines be tan ΞΈ =|(π‘š_2 βˆ’ π‘š_1)/(1 + π‘š_2 π‘š_1 )| Here m1 = Slope of one line = 2 ΞΈ = 60Β° (given) We need to find m2 Putting the values tan 60Β° = |(π‘š_2 βˆ’ 2)/(1 + 2 Γ— π‘š_2 )| √3 = |(π‘š_2 βˆ’ 2)/(1 + 2π‘š_2 )| |(π‘š_2 βˆ’ 2)/(1 + 2π‘š_2 )|= √3 (π‘š_2 βˆ’ 2)/(1 + 2π‘š_2 ) = Β± √3 So, (π‘š_2 βˆ’ 2)/(1 + 2π‘š_2 ) = √3 and (π‘š_2 βˆ’ 2)/(1 + 2π‘š_2 ) = – √3 Taking (π’Ž_𝟐 βˆ’ 𝟐)/(𝟏 + πŸπ’Ž_𝟐 ) = βˆšπŸ‘ m2 βˆ’ 2 = √3(1 + 2m2) m2 βˆ’ 2 = √3 + 2√3m2 m2 – 2√3m2 = √3 + 2 m2 (1 – 2√3) = 2 + √3 m2 = (2 + √3)/(1 βˆ’ 2√3) Taking (π’Ž_𝟐 βˆ’ 𝟐)/(𝟏 + πŸπ’Ž_𝟐 ) = βˆ’ βˆšπŸ‘ m2 βˆ’ 2 = βˆ’ √3(1 + 2m2) m2 βˆ’ 2 = βˆ’ √3 βˆ’ 2√3m2 m2 + 2√3m2 = βˆ’βˆš3 + 2 m2 (1 + 2√3) = 2 βˆ’ √3 m2 = (2 βˆ’ √3)/(1 + 2√3) We know that equation of a line passing through (x1, y1) & having slope m is (y βˆ’ y1) = m(x βˆ’ x1) Equation of a line passing through (2, 3) & having slope (𝟐 + βˆšπŸ‘)/(𝟏 βˆ’ πŸβˆšπŸ‘) is (y βˆ’ 3) = ( (2 + √3))/(1 βˆ’ 2√3) (x βˆ’ 2) (1 βˆ’ 2√3)(y βˆ’ 3) = (2 + √3)(x βˆ’ 2) 1(y βˆ’ 3) βˆ’ 2√3(y βˆ’ 3) = 2(x – 2) + √3(x βˆ’ 2) y – 3 βˆ’ 2√3y + 6√3 = 2x – 4 + √3x βˆ’ 2√3 y βˆ’ 2√3y βˆ’ 4x βˆ’ √3x = – 6√3 – 2√3 – 4 + 3 y (1 βˆ’ 2√3) βˆ’ x(√3 + 2) = – 1 – 8√3 1 + 8√3 = x(√3 + 2) + y (2√3 – 1) (βˆšπŸ‘ + 2)x + (2βˆšπŸ‘ – 1)y = 1 + 8βˆšπŸ‘ Equation of a line passing through (2, 3) & having slope (𝟐 βˆ’ βˆšπŸ‘)/(𝟏 + πŸβˆšπŸ‘) is (y βˆ’ 3) = (2 βˆ’ √3)/(2√3 + 1)(x βˆ’ 2) (2√3 + 1) (y βˆ’ 3) = (2 βˆ’ √3) (x βˆ’ 2) 2√3 (y βˆ’ 3) + 1 (y βˆ’ 3) = 2(x βˆ’ 2) βˆ’ √3(x βˆ’ 2) 2√3 (y βˆ’ 3) + 1 (y βˆ’ 3) = 2(x βˆ’ 2) βˆ’ √3(x βˆ’ 2) 2√3y βˆ’ 6√3 + y βˆ’ 3 = 2x βˆ’ 4 βˆ’ √3x + 2√3 2√3y + y βˆ’ 6√3 βˆ’ 3 = 2x βˆ’ √3x βˆ’ 4 + 2√3 (2√3 + 1)y βˆ’ 6√3 βˆ’ 3 = (2 βˆ’ √3)x βˆ’ 4 + 2√3 (2√3 + 1)y = βˆ’ (√3 βˆ’ 2)x βˆ’ 4 + 2√3 + 6√3 + 3 (2√3 + 1)y + (√3βˆ’2)x = 2√3 + 6√3 βˆ’ 4 + 3 (βˆšπŸ‘βˆ’πŸ)x + (2βˆšπŸ‘ + 1)y = 8βˆšπŸ‘ βˆ’ 1 Hence the equation of lines is (√3 + 2)x + (2√3 – 1)y = 1 + 8√3 or (√3 βˆ’ 2)x + (2√3 + 1)y = 8√3 βˆ’ 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.