Check sibling questions

Ex 10.2, 4 - Find equation (2, 2 root3), is inclined at 75

Ex 10.2, 4 - Chapter 10 Class 11 Straight Lines - Part 2
Ex 10.2, 4 - Chapter 10 Class 11 Straight Lines - Part 3 Ex 10.2, 4 - Chapter 10 Class 11 Straight Lines - Part 4


Transcript

Ex 10.2, 4 Find the equation of the line which passes though (2, 2√3) and is inclined with the x-axis at an angle of 75°. We know that equation of line passing through point (x0, y0) with slope m is y – y0 = m(x – x0) Here Point (x0, y0) = (2, 2√3) Hence x0 = 2, y0 = 2√3 And slope = m = tan θ Given θ = 75° ∴ m = tan(75°) = tan (45 + 30)° = tan⁡〖45° + 〖 tan〗⁡〖30°〗 〗/(1 − tan⁡〖45°tan⁡〖30°〗 〗 ) = (1 + 1/√3)/(1 − 1/√3) = ((√3 + 1)/√3)/((√3 − 1)/√3) = (√3 + 1)/√3 × √3/(√3 − 1) = (√3 + 1)/(√3 − 1) ∴ m = (√3 + 1)/(√(3 ) − 1) ("Using " 𝑡𝑎𝑛⁡〖(𝐴+𝐵)= 𝑡𝑎𝑛⁡〖𝐴 + 𝑡𝑎𝑛⁡𝐵 〗/(1 − 𝑡𝑎𝑛⁡〖𝐴 𝑡𝑎𝑛⁡𝐵 〗 )〗 ) Putting values in (y – y0) = m (x – x0) (y – 2√3) = (√3 + 1)/(√3 − 1) (x – 2) (y – 2√3) (√3 − 1) = (√3 + 1) (x – 2) y (√3 −1) – 2√3 (√3 − 1) = x(√3 + 1) – 2 (√3 + 1) y (√3 − 1) – 2√3 × √3 + 2√3 = x(√3 + 1) – 2√3 – 2 y (√3 − 1) – 6 + 2√3 = x(√3 + 1) – 2√3 – 2 y (√3 − 1) = x(√3 + 1) – 2√3 – 2 + 6 – 2√3 y (√3 − 1) = x(√3 + 1) – 4√3 + 4 y (√3 − 1) – x (√3 + 1) = –4√3 + 4 y (√3 − 1) – x (√3 + 1) = 4(–√3 + 1) x (√𝟑 + 1) – y(√𝟑 – 1) = 4(√𝟑 − 1)

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.