Writing the equation of a perpendicular bisector Write the equation of the perpendicular bisector of the line segment between the points $(1,-2)$ and $( -1,-2)$.
What I have worked out so far:
The first part is $m = \dfrac{y_2-y_1}{x_2-x_1}$
$$m = \frac{-2-(-2)}{-1-1} = \frac{0}{-2}$$
$$m_{perpendicular} =  \frac{2}{0}$$  
Is $\frac{2}{0}$ correct?  (But I can't divide by zero??)  Where do I go from here?

I'll approach this as a more general problem,
using techniques that are probably beyond what this question was
expecting the solution to use. Consider this an answer for future reference.

Problem:
Given $A=(x_A,y_A)$ and $B=(x_B,y_B),$ find an equation of the
perpendicular bisector of the segment $AB.$
Solution:
Define vectors 
$\newcommand{\a}{\mathbf a}\a$ and $\newcommand{\b}{\mathbf b}\b$ 
equal to the displacements of $A$ and $B$ from the origin:
$$
\a = \begin{pmatrix} x_A \\ y_A \end{pmatrix}, \qquad
\b = \begin{pmatrix} x_B \\ y_B \end{pmatrix}.
$$
Now let $\newcommand{\v}{\mathbf v}\v = \b - \a.$
Then the equation 
$$
\v \cdot \begin{pmatrix} x \\ y \end{pmatrix} = c,  \tag1
$$
where $\v\cdot\mathbf u$ is the inner product (vector "dot" product)
of $\v$ and $\mathbf u$,
is the equation of a line perpendicular to $\v,$
and therefore perpendicular to the segment $AB.$
The constant $c$ determines which member of that family of lines
the equation describes.
We want a line perpendicular to $AB$ that passes through the midpoint
of the segment $AB,$ that is, the point 
$\newcommand{\xC}{\frac{x_A+x_B}{2}}\newcommand{\yC}{\frac{y_A+y_B}{2}}
\left(\xC, \yC\right).$
In order for this point to be on the line described by Equation $1,$
it must be true that
$$
\v \cdot \begin{pmatrix}\xC \\ \yC\end{pmatrix} = c.
$$
We can use this fact to substitute for $c$ in Equation $1,$ with the result
$$
\v \cdot \begin{pmatrix} x \\ y \end{pmatrix} =
\v \cdot \begin{pmatrix}\xC \\ \yC\end{pmatrix}.
$$
This is a perfectly valid equation of the perpendicular bisector of segment $AB.$ In a question such as posed here, however, no doubt a "simpler"
form of the equation is desired. 
Such a form can be obtained using the fact that
$$
\v = \begin{pmatrix} x_B - x_A \\ y_B - y_A \end{pmatrix}.
$$
Making this substitution for $\v,$ multiplying term-by-term to evaluate
the inner product, and simplifying algebraically, we have
\begin{align}
\begin{pmatrix} x_B - x_A \\ y_B - y_A \end{pmatrix}
  \cdot \begin{pmatrix} x \\ y \end{pmatrix}
 &= \begin{pmatrix} x_B - x_A \\ y_B - y_A \end{pmatrix}
      \cdot \begin{pmatrix}\xC \\ \yC\end{pmatrix},\\
(x_B - x_A)x + (y_B - y_A)y
 &= (x_B - x_A)\left(\xC\right) + (y_B - y_A)\left(\yC\right),\\
(x_B - x_A)x + (y_B - y_A)y &= \frac12(x_B^2 - x_A^2 + y_B^2 - y_A^2).
\end{align}
For any given points $A$ and $B$ we can then substitute the given values
of $x_A,$ $y_A,$ $x_B,$ and $y_B$ into the equation to obtain the
equation of a line in the simple format $px + qy = k$ for constants
$p,$ $q,$ and $k.$
The advantage of this approach over most others is that it has no special
cases to watch out for (such as when the segment $AB$ is horizontal
or vertical); it works exactly the same for every pair of points
$A$ and $B,$ requiring only that they be distinct points.
But if you must have the equation in a different format
(such as, "$y=mx+b$ unless the line is vertical, in which case write $x=k$"),
it is easy to convert the equation above into the desired format.
In the particular instance given in the question,
$x_A=1,$ $y_A=-2,$ $x_B=-1,$ and $y_B=-2,$
and the equation of the line simplifies to
$$
-2x + 0y = 0,
$$
or even more simply,
$$
x = 0.
$$
