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Example 9 - Chapter 9 Class 11 Straight Lines
Last updated at August 17, 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
- 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 9 Find the distance of the point (3, โ5) from the line 3x โ 4y โ26 = 0. We know that distance (d) of a point (x1, y1) from a line Ax + By + C = 0 is d = |๐ด๐ฅ_1 + ใ๐ต๐ฆใ_2 + ๐ถ|/โ(๐ด^2 + ๐ต^2 ) Now, our equation is 3x โ 4y โ 26 = 0 The above equation is of the form Ax + By + C = 0 where A = 3, B = โ4 , C = โ26 & we have to find the distance of the point (3, โ 5) from the line So, x1 = 3 , y1 = โ5 Now finding distance d = |๐ด๐ฅ_1 + ใ๐ต๐ฆใ_2 + ๐ถ|/โ(๐ด^2 + ๐ต^2 ) Putting values = |3(3) + (โ4)( โ5) โ 26|/โ(32 + (โ4)2) = |9 + 20 โ 26|/โ(9 + 16) = |29 โ 26|/โ25 = |3|/โ(5 ร 5) = |3|/5 = 3/5 โด Required distance = d = ๐/๐ units
