Ex 9.2

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Question 1 Important Deleted for CBSE Board 2024 Exams

Last updated at April 16, 2024 by Teachoo

Ex 9.2, 13 Find equation of the line through the point (0, 2) making an angle 2𝜋/3 with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin. Let AB be the line passing through P(0, 2) & making an angle 2𝜋/3 with positive x-axis Slope of line AB = tan θ = tan (2𝜋/3) = tan (120°) = tan (180 – 60° ) = –tan (60°) = –√3 (tan (180 – θ) = –tan θ) (tan 60° = √3 ) We know that Equation of line passing through (x0, y0) & having slope m (y – y0) = m (x – x0) Equation of line AB passing through (0, 2) & having slope −√3 (y – 2) = –√3(x – 0) y – 2= –√3x y + √3x = 0 + 2 √3x + y = 2 √𝟑x + y − 2 = 0 Hence, equation of line AB is √3x + y − 2 = 0 Also, we have to find equation of line which is parallel to line AB & crossing at a distance of 2 unit below the origin Let CD be the line parallel to AB & passing through point R(0, –2) We know that if two lines are parallel their slopes are equal Therefore, Slope of CD = Slope of AB Slope of CD = –√3 Now Equation of line passing through point (x0, y0) & having slope m (y – y0) = m (x – x0) Equation of line CD passing through (0, -2) & slope –√3 (y – (−2)) = –√3 (x – 0) (y + 2) = √3 (x) (y + 2) = –√3 x y + √3 x + 2 = 0 √𝟑 𝒙 + y + 2 = 0 Hence equation of line CD = √3 𝑥 + y + 2 = 0