You are here
Question 6 - 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
Question 6 Reduce the equation √3x + y − 8 = 0 into normal form. Find the values of p and ω. √3x + y − 8 = 0 " " √3 "x + y" = 8 Dividing by √((√3)2 + (1)2) = √(3 + 1) = √4 = 2 (√3 𝑥)/2 + 𝑦/2 = 8/2 (√3 𝑥)/2 + 𝑦/2 = 4 𝑥(√3/2) + 𝑦(1/2) = 4 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 Normal form of any line is x cos 𝜔 + y sin 𝜔 = p Comparing (1) & (2) p = 4 & cos 𝜔 = √3/2 & sin 𝜔 = 1/2 We know that cos 30° = √3/2 and sin 30° = 1/2 Thus, 𝜔 = 30° So, 𝜔 = 30° & p = 4 Thus, the normal form of line is x cos 30° + y sin 30° = 4
