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Ex 9.3, 10 - Chapter 9 Class 11 Straight Lines
Last updated at August 8, 2026 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
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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
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Transcript
Ex 9.3, 11 Prove that the line through the point (x1, y1) and parallel to the line Ax + By + C = 0 is A (x – x1) + B (y – y1) = 0. The line passing through (x1, y1) and parallel to the line Ax + By + C = 0 has the same slope as the line Ax + By + C = 0 Finding slope of Ax + By + C = 0 By = − Ax − C y = ( − 𝐴𝑥 − 𝐶)/𝐵 y = (( − 𝐴)/𝐵) x + (( −𝐶)/𝐵) The above equation is of the form y = mx + c Where m = slope of line Slope of line (Ax + By + C = 0) = ( − 𝐴)/𝐵 Since both lines are parallel Slope of the required line also same Hence, Slope of required line = − 𝐴/𝐵 Now, finding equation of line We know that equation of a line passing through the point (x1, y1) & having slope m is (y – y1) = m (x – x1) So, equation of line passing through point (x1, y1) & having slope (−𝐴)/𝐵 is (y – y1) = − 𝐴/𝐵(x – x1) B (y – y1) = − A (x – x1) By − By1 = − Ax + Ax1 Ax − Ax1 + By − By1 = 0 A(x – x1) + B(y – y1) = 0 Which is the required equation Hence proved
