# How do you write an equation of a line that goes through ((-6,-2) with slope -1/3?

May 1, 2015

I think it's educational to solve this problem in a general form.
Given a slope $p$ and a point a line passes $\left(a , b\right)$, find its equation.

A line with a slope $p$ can be described by an equation $y = p \cdot x + q$. We do know a coefficient $p$ since this is a given slope, but we do not know a coefficient $q$.

Since we know a point $\left(a , b\right)$ our line passes, the line equation should be satisfied if we substitute $x = a$ and $y = b$ getting an equation for an unknown coefficient $q$:
$b = p \cdot a + q$.

From this equation we derive a value of $q$:
$q = b - p \cdot a$.

So, the equation of a line looks like this:
$y = p \cdot x + b - p \cdot a$

In our particular case:
$p = - \frac{1}{3}$
$a = - 6$
$b = - 2$

So, the equation of a line is
$y = - \frac{1}{3} \cdot x + \left(- 2\right) - \left(- \frac{1}{3}\right) \cdot \left(- 6\right)$
or
$y = - \frac{1}{3} \cdot x - 4$

Here is the graph of this line:
graph{(-1/3)x-4 [-10, 10, -5, 5]}