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Ex 9.3, 3 - 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
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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, 3 Find the distance of the point (ā1, 1) from the line 12(x + 6) = 5(y ā 2). The distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is d = |š“š„_1 + šµš¦_1 + š¶|/ā(š“^2 + šµ^2 ) The given line is 12(x + 6) = 5(y ā 2) 12x + 12 Ć 6 = 5y ā 5 Ć 2 12x + 72 = 5y ā 10 12x ā 5y + 82 = 0 The above equation is of the form Ax + By + C = 0 where A = 12, B = ā5, and C = 82 Now We have to find distance of a point (ā1, 1) from a line So, x1 = ā1 & y1 = 1 Finding distance d = |š“š„_1 + šµš¦_1 + š¶|/ā(š“^2 + šµ^2 ) Putting values d = | ā 12 ā 5 + 82|/ā((12)2 + ( ā 5)2) d = | ā 12 ā 5 + 82|/ā(144 + 25) d = 65/ā169 d = 65/ā(13 Ć 13) d = 65/13 d = 5 Thus, Required distance = 5 units
