Question 703158
<pre>
We locate the point(-4,-2)

{{{drawing(400,400,-6,3,-6,3,

grid(1), circle(-4,-2,.1),circle(-4,-2,.05) )}}}

Then we consider the slope to be {{{(-2)/3}}},
with the numerator negative and the denominator
positive.

The numerator of the slope, -2, is the vertical change. So we
draw a line downward 2 units (downward because
it's negative)

{{{drawing(400,400,-6,3,-6,3, red(line(-4,-2,-4,-4),
line(-3.97,-2,-3.97,-4),line(-4.05,-2,-4.05,-4)
),

grid(1), circle(-4,-2,.1),circle(-4,-2,.05) )}}}

The denominator of the slope, 3, is the horizontal change. So we
draw a line to the right 3 units (rightward because
it's positive)

{{{drawing(400,400,-6,3,-6,3, red(line(-4,-2,-4,-4),
line(-3.97,-2,-3.97,-4),line(-4.05,-2,-4.05,-4),

line(-4,-4,-1,-4),
line(-4,-3.97,-1,-3.97),line(-4,-4.05,-1,-4.05)

),

grid(1), circle(-4,-2,.1),circle(-4,-2,.05) )}}}

Now we just draw a line across the two ends of that 
L-shaped thingy, and that's it:

{{{drawing(400,400,-6,3,-6,3, red(line(-4,-2,-4,-4),
line(-3.97,-2,-3.97,-4),line(-4.05,-2,-4.05,-4),

line(-4,-4,-1,-4),
line(-4,-3.97,-1,-3.97),line(-4,-4.05,-1,-4.05)

), line(-10,2,11,-12),

grid(1), circle(-4,-2,.1),circle(-4,-2,.05) )}}}

Edwin</pre>