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Question
Find the equation of a line passing through (3, −2) and perpendicular to the line x − 3y + 5 = 0.
Solution
The equation of the line perpendicular to x − 3y + 5 = 0 is \[3x + y + \lambda = 0\], where
\[\lambda\] is a constant.
It passes through (3, −2).
\[9 - 2 + \lambda = 0\]
\[ \Rightarrow \lambda = - 7\]
Substituting \[\lambda\] = −7 in \[3x + y + \lambda = 0\], we get
\[3x + y - 7 = 0\] , which is the required line.
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