Find the equation of a line passing through (3, -2) and perpendicular to the line x – 3y + 5 = 0.

Given: equation is perpendicular to x – 3y + 5 = 0 and passes through (3,-2)


To find:


Equation of required line.


Explanation:


The equation of the line perpendicular to x − 3y + 5 = 0 is


3x + y + λ = 0,


Where λ is a constant.


It passes through (3, − 2).


9 – 2 + λ = 0


⇒ λ = - 7


Substituting λ = − 7 in 3x + y + λ = 0,


Hence, we get 3x + y – 7 = 0, which is the required line.


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