# How do you write an equation Perpendicular to the line -3x + y = -7 and Contains the point of: (3,1)?

Jul 22, 2016

$x + 3 y - 6 = 0$.

#### Explanation:

Let $l : - 3 x + y = - 7$, i.e., $y = 3 x - 7$, be the given line, $l '$ the

reqd. line, and the given pt. $A \left(3 , 1\right) \in l '$.

Observe that the slope of $l = 3$, &, as $l ' \bot l ,$the slope of l' has to

be $- \frac{1}{3}$.

Further, $A \left(3 , 1\right) \in l '$.

Hence, using the Slope-Point Form for $l '$, we find,

eqn. of $l ' : y - 1 = - \frac{1}{3} \left(x - 3\right) , i . e . , 3 y - 3 = - x + 3 , \mathmr{and} , x + 3 y - 6 = 0$.