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Reflecting a point over a line

It's astonishing how difficult it is to find a good explanation how to reflect a point over a line that does not use higher math methods. So here is my explanation:

You have a point \(P = (x,y)\) and a line \(g(x) = m \cdot x + t\) and you want to get the point \(P' = (x', y')\) that got mirrored over \(g\).

Reflection point over a line
Reflection point over a line

As you can see, you can construct this quite easily on paper:

  1. Construct the perpendicular through \(P\) to \(g\). It starts in \(L\) and ends in \(P\).
  2. Double the length of the perpendicular in the direction of \(L\).
  3. The endpoint is \(P'\).

How can you do that without drawing it?

First you have to get the perpendicular \(s(x) = m_s \cdot x + t\) (the dashed red line).

You have to know this: \(m_s = - \frac{1}{m}\) And then you know that \(P\) is on \(s\). So you simply put in the values \(x,y\) of P and solve to \(t\): \(t = y - m_s \cdot x\)

Now you have \(s\). As \(s\) and \(g\) have exactly point in common, the following equation gives exactly one result:

\(s(x) = g(x)\)

You have to solve for \(x\). Then you only need to put \(x\) into \(s(x)\) or \(g(x)\) and you're done. You've calculated \(L = (x,y)\).

Now you know \(\Delta x = |x_L - x_P|\) and \(\Delta y = |y_L - y_P|\) and you can calculate \(P'\)


Published

Dez 2, 2012
by Martin Thoma

Category

My bits and bytes

Tags

  • Geometry 5
  • mathematics 59

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