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?
 A: While solving problems on coordinate geometry, it often helps to draw the diagram. If you draw the diagram, you'll notice that the perpendicular bisector of the given line is the y-axis, i.e. $x = 0$.
For a rigorous solution, see the following:
We use the section formula to find out the mid-point of the line segment joining $(1, -2)$ and $(-1, -2)$. Thus the midpoint is $\left ( \frac{1 + (-1)}{2}, \frac{-2 + (-2)}{2} \right ) = (0, -2)$. Now this line segment has a slope of $\tan 0 = 0$. So it's perpendicular bisector must have a slope of $\tan 90°$ meaning it is vertical. So, write the equation of the line as:
$$
(x - 0) = (\cot 90°)(y - (-2)) \\
\implies x = 0
$$
Done.
A: 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,
$$
x = 0.
$$
A: the line through the points $P_1(1,-2)$ and $P_2(-1,-2)$ has the equation $y=-2$ and the perpendicular bisector of these two points is the $y$-axes.
A: Notice that the line passing through $(-1,-2)$ and $(1,-2)$ is parallel to x-axis. So, the perpendicular bisector is parallel to y-axis.$\implies x=c$
And it passes through $(0,-2)$[midpoint]
Hence the perpendicular bisector equation is $x=0$
A: Since you know coordinates of two points of the line $(-1,-2), (1,-2)$, assume that equation the line is of the form $y=mx+c$.
Now putting $(x,y)=(1,-2),(-1,-2)$, will give you the value of $m$ as $m=\frac{-2-(-2)}{2}=0$ 
Let, $m'$ be the slope of perpendicular bisector, then $m\times m'=-1$ or $m'=$ not defined.
Finding the coordinates of mid points of the line as $x=\frac{1-1}{2}=0$ and $y=\frac{-2-2}{2}=-2$ or $(x,y)=(0,-2)$.
Now, you have slope for the perpendicular and also coordinates of one of its points. 
So, $\frac{y-y_1}{x-x_1}=m'\implies \frac{y+2}{x-0}=$ not defined $\implies x=0$.
So, equation of perpendicular bisector is $x=0$ or the perpendicular bisector of the given line is $y-$axis.
Here is graph of your problem:
A: The perpendicular bisector of $A$ and $B$ consists of all points an equal squared-distance away from $A$ and $B$, so:
$$ (x-1)^2 + {(y+2)^2} = (x+1)^2 + {(y+2)^2}$$
$$ \iff x = 0$$
A: To find the required equation follow procedure mentioned below.
1. Find Mid-Point of the given two points. ;
2. Mid-Point is (0,-2). Given line is parallel to the x-axis therefore, any line perpendicular to it will be of the form x = k where k is constant. Because value of abscissa must be constant for a line to be parallel to y-axis and hence, perpendicular to x-axis.
3. In this case k = 0
Therefore,  x = 0 is the required equation.
A: Your approach to link to slope is incorrect.
Equal distance property is to be used.
$$ \sqrt{(x-1)^2 + (y+2)^2}  =  \sqrt{(x+1)^2 + (y+2)^2} $$
Simplifying you get $x=0$- or $y-$ axis.
To verify it, note that between the two points only $x$ coordinate sign has changed. So the locus is halfway in between, that is, the $y-$ axis.
A: A couple of important details:

*

*A line does not have a perpendicular bisector, because a line has no midpoint.  Presumably you actually want to find the perpendcular bisector of the segment that has endpoints at $(1,-2)$ and $(-1, -2)$.  In what follows, I assume that is what the question actually means.

*You seem to be using the fact that the slopes of perpendicular lines are "opposite reciprocals".  A lot of students (and even a lot of teachers!) forget  this is only true for lines that are neither vertical nor horizontal.  If a line is vertical, then any line perpendicular to it is horizontal, and vice versa.

So now let's figure out what the segment going from  $(1,-2)$ to $(-1,-2)$ actually looks like.  The two points have the same $y$-coordinate, which means the segment is horizontal; see below.

You can probably tell directly from the graph that the midpoint is at $(0,-2)$, but if that isn't obvious (or you want something more computational to be convinced) you can use the midpoint formula.
So now you want a vertical line that goes through $(0,-2)$.  What line is that?  It's the line made up of all points whose $x$-coordinate is $0$.  In other words, the equation of the perpendicular bisector is $x=0$ -- which is actually the $y$-axis itself.
