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Example 9 - 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
- 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
Example 9 Find the distance of the point (3, ā5) from the line 3x ā 4y ā26 = 0. We know that distance (d) of a point (x1, y1) from a line Ax + By + C = 0 is d = |š“š„_1 + ćšµš¦ć_2 + š¶|/ā(š“^2 + šµ^2 ) Now, our equation is 3x ā 4y ā 26 = 0 The above equation is of the form Ax + By + C = 0 where A = 3, B = ā4 , C = ā26 & we have to find the distance of the point (3, ā 5) from the line So, x1 = 3 , y1 = ā5 Now finding distance d = |š“š„_1 + ćšµš¦ć_2 + š¶|/ā(š“^2 + šµ^2 ) Putting values = |3(3) + (ā4)( ā5) ā 26|/ā(32 + (ā4)2) = |9 + 20 ā 26|/ā(9 + 16) = |29 ā 26|/ā25 = |3|/ā(5 Ć 5) = |3|/5 = 3/5 ā“ Required distance = d = š/š units
