Ex 9.2, 13 - Line through (0, 2) making angle 2pi/3 - Ex 9.2 - Ex 9.2

part 2 - Ex 9.2, 13 - Ex 9.2 - Serial order wise - Chapter 9 Class 11 Straight Lines
part 3 - Ex 9.2, 13 - Ex 9.2 - Serial order wise - Chapter 9 Class 11 Straight Lines part 4 - Ex 9.2, 13 - Ex 9.2 - Serial order wise - Chapter 9 Class 11 Straight Lines

 

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Ex 9.2, 13 Find equation of the line through the point (0, 2) making an angle 2πœ‹/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œ‹/3 with positive x-axis Slope of line AB = tan ΞΈ = tan (2πœ‹/3) = tan (120Β°) = tan (180 – 60Β° ) = –tan (60Β°) = β€“βˆš3 (tan (180 – ΞΈ) = –tan ΞΈ) (tan 60Β° = √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 βˆ’βˆš3 (y – 2) = β€“βˆš3(x – 0) y – 2= β€“βˆš3x y + √3x = 0 + 2 √3x + y = 2 βˆšπŸ‘x + y βˆ’ 2 = 0 Hence, equation of line AB is √3x + y βˆ’ 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 = β€“βˆš3 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 β€“βˆš3 (y – (βˆ’2)) = β€“βˆš3 (x – 0) (y + 2) = √3 (x) (y + 2) = β€“βˆš3 x y + √3 x + 2 = 0 βˆšπŸ‘ 𝒙 + y + 2 = 0 Hence equation of line CD = √3 π‘₯ + y + 2 = 0

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