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Question 1 - Normal form - Chapter 9 Class 11 Straight Lines
Last updated at Dec. 16, 2024 by Teachoo
Chapter 9 Class 11 Straight Lines
Concept wise
- Cordinate geometry questions
- Slope - Finding slope
- Slope - Prependicularity
- Angle between two lines by Slope
- Collinearity of 3 points by sliope
- Point Slope form
- Slope-Intercept form
- Intercept form
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Normal form
- Given Statement
- Two lines // or/and prependicular
- Angle between two lines
- Distance - Between two parallel lines
- Distance of a point from a line
- Distance of a point from a line along a line
- Other Type of questions - Area of Triangle formed
- Other Type of questions - 3 lines Concurrent
- Other Type of questions - Image
- Other Type of questions - Mix
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
