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Misc 2 - Find values of theta, p, if equation is normal form

Misc 2 - Chapter 10 Class 11 Straight Lines - Part 2
Misc 2 - Chapter 10 Class 11 Straight Lines - Part 3

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Misc 2 Find the values of πœƒ and p, if the equation is the normal form of the line √3x + y + 2 = 0 . √3x + y + 2 = 0 2 = – √3x – y β€“βˆš3x – y = 2 Dividing by √((βˆ’βˆš3)2 + (βˆ’1)2) = √(3+1) = √4 = 2 both sides (βˆ’βˆš3 π‘₯)/2 βˆ’ 𝑦/2 = 2/2 (βˆ’βˆš3 π‘₯)/2 βˆ’ 𝑦/2 = 1 Normal form is x cos πœ” + y sin πœ” = p Where p is the perpendicular distance from origin & πœ” is the angle between perpendicular & the positive x-axis ((βˆ’βˆš3)/2)π‘₯ + ((βˆ’1)/2)y = 1 Normal form of any line is x cos πœ” + y sin πœ” = p Comparing (1) & (2)a p = 1 & cos Ο‰ = (βˆ’βˆš3)/2 & sin Ο‰ = (βˆ’1)/2 Now, finding Ο‰ ∴ Ο‰ = 180Β° + 30Β° = 210Β° Rough Ignoring signs cos ΞΈ = √3/2 & sin ΞΈ = 1/2 So, ΞΈ = 30Β° As sin & cos both are negative, ∴ 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œ‹/(180Β° ) = πŸ•π…/πŸ” & Perpendicular Distance = p = 1

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.