# How do you write an equation of a line passing through (6, 1), perpendicular to 3x + y = 7?

##### 1 Answer
Jul 18, 2016

$x - 3 y = 3$

#### Explanation:

Any line perpendicular to $3 x + y = 7$ must be of the form

$x - 3 y = c$

where $c$ is a constant (this is easy to see from the fact that the slopes ${m}_{1}$ and ${m}_{2}$ of two mutually perpendicular lines obey ${m}_{1} {m}_{2} = - 1$). Since the point $\left(6 , 1\right)$ lies on this line, the constant $c$ is given by

$c = 6 - 3 \times 1 = 3$

The straight line that we seek is

$x - 3 y = 3$