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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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.