Equation of the line, vertical I need to find the equation of the line, so I want to use y-y1=m(x-x1)  but for a vertical line the slope is undefined.  So I am unsure how to proceed.  
 A: A vertical line in the plane is defined by the equation $x = c$, where $c$ is some real. This makes sense because along the vertical line, the $x$ value doesn't change. 
A: Here's a hint:  there is one thing in common among the coordinates of all of the points on a vertical line.  Express this common feature as an equation an you'll have the equation of your line.
A: You can also think of it using the slightly massaged one:
dx*(y-b) = dy*(x-a)
for the line passing through (a,b) and "having slope" dy/dx  (here dx = 0 )
A: use the equation of a line: $ax+by+c=0$
A: Seeing as this is tagged as linear algebra:
You might be interested in expressing your line using a parametric equation involving vectors.
For example, if you want a vertical line (in the direction of the standard basis vector $(0,1)$) which passes through the point $(3,7)$, then you could use
$$(x,y)=(3,7)+t\cdot (0,1)$$
where $t$ varies over all scalars (elements of $\mathbb{R}$ probably).
In this particular case of a vertical line, as others have noted, the equation would be equivalent to just writing $$x=3+t\cdot 0=3$$ since the $x$ value does not vary, and the $y$ value is not constrained.
More generally, the equation for the line in the direction of vector $(\Delta x,\Delta y)$ and passing through the point $(x_0,y_0)$ would be
$$(x,y) = (x_0,y_0) + t\cdot (\Delta x,\Delta y)$$
You should understand this as an equation which describes the collection of points which includes $(x_0, y_0)$, along with any point we get by starting at $(x_0, y_0)$ and moving some distance (forward or backward) in the direction of $(\Delta x, \Delta y)$.  If you think about this a bit, you should see that this has the advantage of generalizing fairly easily to higher dimensional spaces, where you've got more variables than just $x$ and $y$.
A: How about below parameterization.
$ \rho = x \cos{\theta} + y \sin{\theta}$ 
where 
$ \rho $ = perpendicular distance to the line from the origin (0,0)
$ \theta $ = angle of a line perpendicular to our line with the x axis
you can take a look at http://en.wikipedia.org/wiki/Hough_transform.
when line slope is $90^o$, $\theta = 0^o$, then $ \rho = x $.
