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

1 Answer
Zor Shekhtman
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 #(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]}

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