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Last updated at Feb. 3, 2020 by Teachoo

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Misc 2 Find the values of ๐ and p, if the equation is the normal form of the line โ3x + y + 2 = 0 . โ3x + y + 2 = 0 2 = โ โ3x โ y โโ3x โ y = 2 Dividing by โ((โโ3)2 + (โ1)2) = โ(3+1) = โ4 = 2 both sides (โโ3 ๐ฅ)/2 โ ๐ฆ/2 = 2/2 (โโ3 ๐ฅ)/2 โ ๐ฆ/2 = 1 Normal form is x cos ๐ + y sin ๐ = p Where p is the perpendicular distance from origin & ๐ is the angle between perpendicular & the positive x-axis ((โโ3)/2)๐ฅ + ((โ1)/2)y = 1 Normal form of any line is x cos ๐ + y sin ๐ = p Comparing (1) & (2)a p = 1 & cos ฯ = (โโ3)/2 & sin ฯ = (โ1)/2 Now, finding ฯ โด ฯ = 180ยฐ + 30ยฐ = 210ยฐ Rough Ignoring signs cos ฮธ = โ3/2 & sin ฮธ = 1/2 So, ฮธ = 30ยฐ As sin & cos both are negative, โด ฯ will lie in 3rd quadrant, So, ฯ = 180ยฐ + 30ยฐ So, the normal form of line is x cos 210ยฐ + y sin 210ยฐ = 1 Hence, Angle = 210ยฐ = 210 ร ๐/(180ยฐ ) = ๐๐ /๐ & Perpendicular Distance = p = 1

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.