Equation of a line that passes halfway between two points (in other words, divides the space) Is there a formal proper way of finding the line between two points?
By that I don't mean the line connecting the two points, I mean a line that runs the same distance away from point 1 and point 2.
To phrase it another way, I want to find the equation of a line that divides the plane into two equal parts, where each of the two points are the same distance from the line.
I drew a picture. In this picture, how do I find the purple line?

It may or not be relevant, but I'm asking because I am trying to learn about Support Vector Machines.

Let $(x_1,y_1)$ be the coordinates of point $x_1$ and $(x_2,y_2)$ be the coordinates of point $x_2$. 
The gradient $m$ of the line connecting them will be
$$m=\frac{y_2-y_1}{x_2-x_1}$$
As the purple line will be perpendicular to the line joining points $x_1$ and $x_2$ (because all points on the purple line will be equidistant from points $x_1$ and $x_2$), its gradient $m'$ will be
$$m'=-\frac{1}{m}=\frac{x_1-x_2}{y_2-y_1}$$
Now the point at which the purple line will intersect the line joining $x_1$ and $x_2$ will be the mid-point between $x_1$ and $x_2$:- 
$$\left(x_1+\frac{1}{2}(x_2-x_1),y_1+\frac{m}{2}(x_2-x_1)\right)=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$
We can find the intercept $c$ of the purple line as follows:-
$$\frac{y_1+y_2}{2}=m'\frac{x_1+x_2}{2}+c=\frac{x_1-x_2}{y_2-y_1}\frac{x_1+x_2}{2}+c
\\\Rightarrow c=\frac{(y_2^2-y_1^2)-(x_1^2-x_2^2)}{2(y_2-y_1)}=\frac{(x_2^2+y_2^2)-(x_1^2+y_1^2)}{2(y_2-y_1)}$$
Having found the intercept and the gradient, the equation of the purple line can be expressed as:-
$$y=\frac{x_1-x_2}{y_2-y_1}x+\frac{(x_2^2+y_2^2)-(x_1^2+y_1^2)}{2(y_2-y_1)}$$
