I've been doing some personal math stuff and have spent the last few hours trying to figure this out with no success.

I want to find the coordinates of the point at the end of the small line segment

Here's a picture:


What I know:

  • point1 ( in quad 2 )
  • point2 ( in quad 1 on the edge of the circle )
  • length of the little line

From there I know I can calculate the midpoint, the length of the long line, the measure of the angle formed by connected the first point to the end of the line segment, and all sorts of other things, but I don't know how to use any of it.

I'm trying to whittle away a formula that will at least give the x coordinate of the point in terms of the coordinates of the other points. I think the problem I'm having is that there are two points along the line perpendicular to the longer line that have the same distance and any equation I find is going to include both possibilities.

I'd prefer a hint at how to tackle this then the full-fledged answer, but anything will do at this point.



  • For a vector $(a,b)$ the perpendicular direction is $(-b,a)$ (this transformation turns left, multiply by $-1$ if you want to go the other way).
  • If you want to have a vector along direction $(a,b)$ of length $d$, then calculate $\frac{d}{|(a,b)|}(a,b)$ where $|v|$ is the length of $v$.

Good luck!

  • $\begingroup$ The point wouldn't intersect either (-b,a) or (a,b), though. Am I misunderstanding this? $\endgroup$ – mowwwalker Mar 14 '13 at 7:10
  • 1
    $\begingroup$ @Walkerneo You are, $(a,b) = P_2-P_1$, $M = \frac{P_1+P_2}{2}$, $X = M + \alpha \cdot (-b,a)$ for some $\alpha \in \mathbb{R}$; try calculating $\alpha$ on your own. $\endgroup$ – dtldarek Mar 14 '13 at 7:21
  • $\begingroup$ Thanks so much, it finally snapped with me yesterday after going over that comment 100 times. I appreciate the answer! $\endgroup$ – mowwwalker Mar 14 '13 at 16:29

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