Question

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

Answer 1

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

A line with a slope `p` can be described by an equation `y=p*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 `(a,b)` 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*a+q`.

From this equation we derive a value of `q`:
`q=b-p*a`.

So, the equation of a line looks like this:
`y=p*x+b-p*a`

In our particular case:
`p=-1/3`
`a=-6`
`b=-2`

So, the equation of a line is
`y=-1/3*x+(-2)-(-1/3)*(-6)`
or
`y=-1/3*x-4`

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