Ex 10.2, 8 - Line which is at a perpendicular distance of 5 units - Ex 10.2

  1. Class 11
  2. Important Question for exams Class 11
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Ex10.2, 8 Find the equation of the line which is at a perpendicular distance of 5 units from the origin and the angle made by the perpendicular with the positive x-axis is 30 We need to calculate equation of line Perpendicular distance of line from origin is 5 unites & Normal makes an angle of 30 with the positive x-axis By the normal from Equation of line is x cos + y sin = p. where, p = normal distance from the origin & = angle which makes by the normal with positive x-axis Here p = 5 & = 30 Putting values x cos + y sin = p x cos 30 + y sin 30 = 5 x 3/2 + y 1/2 = 5 ( 3 + )/2 = 5 3 + y = 10 3 + 10 = 0 Thus, equation of line is 3 + 10 = 0

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.