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  1. Chapter 10 Class 11 Straight Lines
  2. Serial order wise

Transcript

Misc 2 Find the values of ๐œƒ and p, if the equation is the normal form of the line โˆš3x + y + 2 = 0 . โˆš3x + y + 2 = 0 2 = โ€“ โˆš3x โ€“ y โ€“โˆš3x โ€“ y = 2 Dividing by โˆš((โˆ’โˆš3)2 + (โˆ’1)2) = โˆš(3+1) = โˆš4 = 2 both sides (โˆ’โˆš3 ๐‘ฅ)/2 โˆ’ ๐‘ฆ/2 = 2/2 (โˆ’โˆš3 ๐‘ฅ)/2 โˆ’ ๐‘ฆ/2 = 1 Normal form is x cos ๐œ” + y sin ๐œ” = p Where p is the perpendicular distance from origin & ๐œ” is the angle between perpendicular & the positive x-axis ((โˆ’โˆš3)/2)๐‘ฅ + ((โˆ’1)/2)y = 1 Normal form of any line is x cos ๐œ” + y sin ๐œ” = p Comparing (1) & (2)a p = 1 & cos ฯ‰ = (โˆ’โˆš3)/2 & sin ฯ‰ = (โˆ’1)/2 Now, finding ฯ‰ โˆด ฯ‰ = 180ยฐ + 30ยฐ = 210ยฐ Rough Ignoring signs cos ฮธ = โˆš3/2 & sin ฮธ = 1/2 So, ฮธ = 30ยฐ As sin & cos both are negative, โˆด 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œ‹/(180ยฐ ) = ๐Ÿ•๐…/๐Ÿ” & Perpendicular Distance = p = 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 9 years. He provides courses for Maths and Science at Teachoo.