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?
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.
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.
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, $$ x = 0. $$
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 $x=0$
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.
The perpendicular bisector of $A$ and $B$ consists of all points an equal squared-distance away from $A$ and $B$, so:
$$ (x-1)^2 + {(y+2)^2} = (x+1)^2 + {(y+2)^2}$$
$$ \iff x = 0$$
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.
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 couple of important details:
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.