Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This problem came up while discussing using a simplex to solve systems of equations.
(By the way, yes, this is very similar to this one.)

Given three points, how do I find the location of the point that results from reflecting one of them over the line between the other two?

This is what I mean:
Whoohoo MS Paint! :P

How do I find $C'$?

share|cite|improve this question
Why are you answering your own question, immediately after it's asked? Perhaps you should go ahead an accept your own answer? – mixedmath May 25 '11 at 15:47
That's interesting. Having read through that topic and the one referenced, I suppose I just thought along the same lines as Pete Clark with respect to questions where the OP knows the answer. Thank you for the explanation. – mixedmath May 25 '11 at 15:55
@mixedmath: You're welcome. :) – El'endia Starman May 25 '11 at 15:58
@mixedmath: look at El'endia's profile... Quite an awesome youthful grandma (or -pa?!) we have here :) – t.b. May 25 '11 at 16:09
up vote 4 down vote accepted

For reference:
WHOOHOO More MS Paint! :P

$\vec{P} = \langle x-a, y-b \rangle$
$\vec{Q} = \langle c-a, d-b \rangle$
$\theta = \text{ the angle between } \vec{P} \text{ and } \vec{Q}$

First off, let's start with projecting $\vec{P}$ onto $\vec{Q}$. In math...

$\vec{K} = \text{Proj}_{\vec{Q}} \vec{P} = \displaystyle \frac{\vec{P} \cdot \vec{Q}}{||\vec{Q}||^2} \vec{Q}$

This gives us the vector $\vec{K}$ that goes from $A$ to the "intersection" of the two lines $\overline{AB}$ and $\overline{CC'}$. To find the vector from $C$ to that intersection, simply subtract $\vec{K}$ from $\vec{P}$. You can then multiply this vector by two and add to $C$ to get $C'$, or in other words $2(\vec{K}-\vec{P}) + \vec{P}$. This is equivalent to $2\vec{K}-\vec{P}$. Substituting the formula for $\vec{K}$ back in gives:

$\vec{P'} = \displaystyle 2\frac{\vec{P} \cdot \vec{Q}}{||\vec{Q}||^2} \vec{Q} - \vec{P}$

You can now add $\vec{P'}$ to $A$ to get $C'$. That sufficient for your needs?

share|cite|improve this answer
Plenty sufficient for mine, in fact. Nice question and answer! – t.b. May 25 '11 at 15:55
@Theo: Thanks! :) – El'endia Starman May 25 '11 at 15:58
By the way: did you ever try GeoGebra? After a short while of getting used to it, you can create quite awesome pictures with that. – t.b. May 25 '11 at 16:00
@Theo: Oh cool! Definitely gonna check that out! :D – El'endia Starman May 25 '11 at 16:03
Maybe it is interesting for you that your approach can be generalized to the reflection at planes in arbitrary dimensions. It is known (at least numerical people) as Householder transformation. – Fabian May 25 '11 at 16:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.