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Question
Find the equation of the line passing through the point (−3, 5) and perpendicular to the line joining (2, 5) and (−3, 6).
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Solution
The given points are \[A \left( 2, 5 \right) \text { and } B \left( - 3, 6 \right)\].
\[\therefore\] Slope of AB \[= \frac{6 - 5}{- 3 - 2} = - \frac{1}{5}\]
Let m be the slope of the required line. Then,
\[m \times \text { Slope of } AB = - 1\]
\[ \Rightarrow m \times \frac{- 1}{5} = - 1\]
\[ \Rightarrow m = 5\]
So, the equation of the line that passes through (−3, 5) and has slope 5 is
\[y - 5 = 5\left( x + 3 \right)\]
\[ \Rightarrow 5x - y + 20 = 0\]
Hence, the equation of the required line is
\[5x - y + 20 = 0\]
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