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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?

Mhenni Benghorbal
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dato datuashvili
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  • This is discussed in the general case at http://math.stackexchange.com/questions/38665/equation-of-angle-bisector-given-the-equations-of-two-lines-in-2d – Mark Bennet Jul 06 '13 at 17:30

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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 $tan(\frac{arctan(m)+arctan(s)}{2}), tan(\frac{arctan(m)+arctan(s)+\pi}{2})$.

Loki Clock
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  • so it means that slope of unknown line is one half one of their slope? – dato datuashvili Jul 06 '13 at 14:58
  • @dato It does not. Your first two slopes were $-2, 2$, so the first new slope is $\frac{-2+2}{2}=0.$ I should have said "slopes halfway between those of the two lines you started from." – Loki Clock Jul 06 '13 at 15:00
  • but second slope is $1/0$ it can't be right? – dato datuashvili Jul 06 '13 at 15:01
  • @dato The first line's slope is 0, which means a line orthogonal to it is vertical, or of infinite slope. For example, try converting your equations into functions of $y$ and applying this method! – Loki Clock Jul 06 '13 at 15:02
  • In your text you have suggested taking the average of the slopes $m$ to get the bisector. It is not as simple as that because the slope $m$ is the tangent of the angle with the axis, and it is the angles you need to average not their tangents. See http://math.stackexchange.com/questions/38665/equation-of-angle-bisector-given-the-equations-of-two-lines-in-2d – Mark Bennet Jul 06 '13 at 17:29
  • @MarkBennet Ack, you're right. It would be so much easier, though. Still, if the slopes are negatives of eachother the bisectors will still be horizontal and vertical, so that's useful to check first. – Loki Clock Jul 06 '13 at 17:57
  • Indeed - the context of this question was unclear, but solving it did not need the full machinery. – Mark Bennet Jul 06 '13 at 18:01
  • Also, if $a$ is the first solution's slope, the second's slope is still $-\frac{1}{a}$, so it's not necessary to compute two separate tangents. – Loki Clock Jul 06 '13 at 18:09
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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 $x=2$.

Hagen von Eitzen
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