Question 495028
Write a standard form of the line that passes through the point (-10,-9), and
 (-10,-7)
<pre>
Notice that when we try to use the slope formula, it fails
because division by 0 is not defined:
{{{ cross(
matrix(4,1,

m=(y[2]-y[1])/(x[2]-x[1]),

m = ((-7)-(-9))/((-10)-(-10)),

m = (-7+9)/(-10+10),

m = 2/0) )}}}

So not only can't use the slope formula, but
we also can't use the slope-intercept formula,
or the point-slope formula because those
involve the slope, and this line doesn't have 
any slope!

So let's plot the two given points:

{{{drawing(400,400,-15,5,-14,6, graph(400,400,-15,5,-14,6),
circle(-10,-7,.1),circle(-10,-9,.1),locate(-10+.4,-7+.4,"(-10,-7)"),
locate(-10+.4,-9+.4,"(-10,-9)") )}}}

Now we draw a line (in green) through them:

{{{drawing(400,400,-15,5,-14,6, graph(400,400,-15,5,-14,6),
circle(-10,-7,.1),circle(-10,-9,.1),locate(-10+.4,-7+.4,"(-10,-7)"),
locate(-10+.4,-9+.4,"(-10,-9)"), green(line(-10,-20,-10,20)) )}}}

We notice that we have a vertical line, the only kind of line
that does not have a slope at all.  Horizontal lines have a slope
of 0 but vertical lines have no slope at all.  Notice the
difference between "having slope 0" and "having no slope at all".

This one has no slope at all.

Let's look at the coordinates of some other points in that line:

{{{drawing(400,400,-15,5,-14,6, graph(400,400,-15,5,-14,6),
circle(-10,-7,.1),circle(-10,-9,.1),locate(-10+.4,-7+.4,"(-10,-7)"),
locate(-10+.4,-9+.4,"(-10,-9)"), green(line(-10,-20,-10,20)),

red(
circle(-10,5,.1), locate(-10+.4,5+.4,"(-10,5)"),
circle(-10,4,.1), locate(-10+.4,4+.4,"(-10,4)"),
circle(-10,3,.1), locate(-10+.4,3+.4,"(-10,3)"),
circle(-10,2,.1), locate(-10+.4,2+.4,"(-10,2)"),
circle(-10,1,.1), locate(-10+.4,1+.4,"(-10,1)"),
circle(-10,0,.1),
circle(-10,-1,.1), locate(-10+.4,-1+.4,"(-10,-1)"),
circle(-10,-2,.1), locate(-10+.4,-2+.4,"(-10,-2)"),
circle(-10,-3,.1), locate(-10+.4,-3+.4,"(-10,-3)"),
circle(-10,-4,.1), locate(-10+.4,-4+.4,"(-10,-4)"),
circle(-10,-5,.1), locate(-10+.4,-5+.4,"(-10,-5)"),
circle(-10,-6,.1), locate(-10+.4,-6+.4,"(-10,-6)"),

circle(-10,-8,.1), locate(-10+.4,-8+.4,"(-10,-8)"),


circle(-10,-10,.1), locate(-10+.4,-10+.4,"(-10,-10)"),
circle(-10,-11,.1), locate(-10+.4,-11+.4,"(-10,-11)"),
circle(-10,-12,.1), locate(-10+.4,-12+.4,"(-10,-12)")    ))}}}

Notice that they all have -10 as their x coordinate.
So we could write:

"x-coordinate = -10"

but instead we take out the "-coordinate" and just write

 x = -10

And that is a standard form of the line.  You may have been told that
a standard form of a line is 

Ax + By = C

x = -10 is really in that form because it is the same as

1x + 0y = -10, where A=1, B=0, and C=-10,

but it is preferable to leave out the 1 coefficient of x
and the term 0y.

So the answer is simply

x = -10

Edwin</pre>