find symmetric line of given two line I have one question. Suppose that we have two lines given by equations

$$y=2x+3$$
  $$y=-2x+11$$

I want to find  all equations of lines which these two given lines have same distances from them in plane.As I know symmetric means that the distance between it and the two given lines must be equal. So suppose  that  the requested line is $$y=kx+b$$
after  calculating distances I found the following equation
$2k-7b=0$    what does it mean? May you help me?
 A: When we say something in a plane is symmetric about a line, we mean it's reflection over that line is unchanged. So this is misleading. How do you define the distance between two lines? The angle? If so, you are looking for an angular bisector of the two lines. But there are two, the vertical line and horizontal line passing through the intersection point of those two lines. The coordinates of this intersection point are calculated as follows:
$$2x+3-y=0, -2x+11-y=0$$
$$\implies 0=(2x+3-y)-(-2x+11-y)=4x-8$$
$$\implies x=2, \implies 2x+3-y=7-y=0 \implies y=7.$$
Thus, the vertical line is given by the graph of x=2, the horizontal by the graph of y=7.
In general, you would take this intersection point and find the lines passing through it whose angle at the intersection is one of the two which are halfway between those of the two lines you started from. Since the slopes $m, s$ of your starting lines are the tangent of these two angles, one of your angles is given by the average of the arctangents of those two slopes, $\frac{arctan(m)+arctan(s)}{2}$, and the other is that angle plus $\frac{\pi}{2}$. Hence the slopes of your solution lines are $tan(\frac{arctan(m)+arctan(s)}{2}), tan(\frac{arctan(m)+arctan(s)+\pi}{2})$.
A: I don't know what you calculated, but for reference, the two lines you are looking for (the angular bisectors) are given by $y=7$ and by $x=2$.
