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 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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.