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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Transcript

Question 1 Find the values of 𝜃 and p, if the equation is the normal form of the line √3x + y + 2 = 0 . √3x + y + 2 = 0 2 = – √3x – y –√3x – y = 2 Dividing by √((−√3)2 + (−1)2) = √(3+1) = √4 = 2 both sides (−√3 𝑥)/2 − 𝑦/2 = 2/2 (−√3 𝑥)/2 − 𝑦/2 = 1 Normal form is x cos 𝜔 + y sin 𝜔 = p Where p is the perpendicular distance from origin & 𝜔 is the angle between perpendicular & the positive x-axis ((−√3)/2)𝑥 + ((−1)/2)y = 1 Normal form of any line is x cos 𝜔 + y sin 𝜔 = p Comparing (1) & (2)a p = 1 & cos ω = (−√3)/2 & sin ω = (−1)/2 Now, finding ω ∴ ω = 180° + 30° = 210° Rough Ignoring signs cos θ = √3/2 & sin θ = 1/2 So, θ = 30° As sin & cos both are negative, ∴ ω will lie in 3rd quadrant, So, ω = 180° + 30° So, the normal form of line is x cos 210° + y sin 210° = 1 Hence, Angle = 210° = 210 × 𝜋/(180° ) = 𝟕𝝅/𝟔 & 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.