Last updated at Feb. 3, 2020 by

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Misc 2 Find the values of π and p, if the equation is the normal form of the line β3x + y + 2 = 0 . β3x + y + 2 = 0 2 = β β3x β y ββ3x β y = 2 Dividing by β((ββ3)2 + (β1)2) = β(3+1) = β4 = 2 both sides (ββ3 π₯)/2 β π¦/2 = 2/2 (ββ3 π₯)/2 β π¦/2 = 1 Normal form is x cos π + y sin π = p Where p is the perpendicular distance from origin & π is the angle between perpendicular & the positive x-axis ((ββ3)/2)π₯ + ((β1)/2)y = 1 Normal form of any line is x cos π + y sin π = p Comparing (1) & (2)a p = 1 & cos Ο = (ββ3)/2 & sin Ο = (β1)/2 Now, finding Ο β΄ Ο = 180Β° + 30Β° = 210Β° Rough Ignoring signs cos ΞΈ = β3/2 & sin ΞΈ = 1/2 So, ΞΈ = 30Β° As sin & cos both are negative, β΄ Ο will lie in 3rd quadrant, So, Ο = 180Β° + 30Β° So, the normal form of line is x cos 210Β° + y sin 210Β° = 1 Hence, Angle = 210Β° = 210 Γ π/(180Β° ) = ππ /π & Perpendicular Distance = p = 1

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Chapter 10 Class 11 Straight Lines (Term 1)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.