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NCERT Class XI Mathematics - Straight Lines - Solutions

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Question : 24 of 74
Marks: +1, -0
Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).
Solution:  
Let M(2, 5) and N(–3, 6) be the end points of the given line segment.
Slope of MN = 6−5−3−2\frac{6-5}{-3-2} = −15\frac{-1}{5}
Since LP ⊥ MN
∴ Slope of LP is 5 and passing through (–3, 5)
Equation of line LP is (y – 5) = 5(x + 3)
⇒ y – 5 = 5x + 15
⇒ 5x – y + 20 = 0.
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