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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 ω + y sin ω = p. where, p = normal distance from the origin & ω = angle which makes by the normal with positive x-axis Here p = 5 & ω = 30° Putting values x cos ω + y sin ω = p x cos 30° + y sin 30° = 5 x √3/2 + y 1/2 = 5 (√3 𝑥 + 𝑦)/2 = 5 √3 𝑥 + y = 10 √3 𝑥 + 𝑦 – 10 = 0 Thus, equation of line is √3 𝑥 + 𝑦 – 10 = 0

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
  • Sharat Singh's image

    Find the equation of the line on which the perpendicular from the origin makes an angele of 30with the x axis and which forms a triangle of area 50/3 with the coordinate axes. 

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