You are here
Question 3 - 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
-
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
Example 11 Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with positive direction of x-axis is 15 . We need to find equation of line Perpendicular distance of AB from origin is 4 units & angle which the normal makes with (+)ve direction of x-axis is 15 By Normal from Equation of line is x cos + y sin = p where, p = normal distance from the origin & = angle which makes by the normal with positive x-axis Here p = 4 & = 15 Putting values x cos + y sin = p x cos 15 + y sin 15 = 4 x (( 3 + 1)/(2 2)) + y (( 3 1)/(2 2)) = 4 ( ( 3 + 1) + ( 3 1))/(2 2) = 4 x( 3 + 1) + y( 3 1) = 4 2 2 ( 3 + 1)x + ( 3 1)y = 8 2 Which Is the required equation
