Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Ex 9.2

Ex 9.2, 1
Important

Ex 9.2, 2

Ex 9.2, 3

Ex 9.2, 4 Important You are here

Ex 9.2, 5

Ex 9.2, 6 Important

Ex 9.2, 7

Ex 9.2, 8 Important

Ex 9.2, 9

Ex 9.2, 10 Important

Ex 9.2, 11

Ex 9.2, 12

Ex 9.2, 13 Important

Ex 9.2, 14 Important

Ex 9.2, 15

Ex 9.2, 16 Important

Ex 9.2, 17 Important

Ex 9.2, 18 Important

Ex 9.2, 19

Question 1 Important Deleted for CBSE Board 2024 Exams

Last updated at May 29, 2023 by Teachoo

Ex 9.2, 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 + 𝑡𝑎𝑛𝐵 〗/(1 − 𝑡𝑎𝑛〖𝐴 𝑡𝑎𝑛𝐵 〗 )〗 ) 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)