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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 equation of the line which is at a perpendicular distance of 5 units from the origin and the angle made by the perpendicular with the positive x-axis is 30° We need to calculate equation of line Perpendicular distance of line from origin is 5 units & Normal makes an angle of 30° with the positive x-axis By the normal from Equation of line is x cos ω + y sin ω = p. where, p = normal distance from the origin & ω = angle which makes by the normal with positive x-axis Here p = 5 & ω = 30° Putting values x cos ω + y sin ω = p x cos 30° + y sin 30° = 5 x √3/2 + y 1/2 = 5 (√3 𝑥 + 𝑦)/2 = 5 √𝟑 𝒙 + y = 10 Thus, equation of line is √3 𝑥 + y = 10
