What is the matrix used to find the reflected (x, y) coordinate in the line y=mx? I hope this makes sense, I'm essentially looking for a matrix in which you can just substitute in the gradient m from y=mx and find the reflected coordinates? If this doesn't make any sense please say why? Regards Tom
 A: The matrix you are searching is:
$$
\dfrac{1}{1+m^2}
\left[
\begin{array}{cccc}
1-m^2&2m\\
2m & m^2-1
\end{array}
\right]
$$
For a proof see may answer to Linear Transformation with eigenvectors and eigenvalues
A: Your question does make sense, since you describe a linear transformation. Any reflection about a line through the origin is a linear transformation.
One way to find the transformation is to find the image of the point $(1,0)$ [call it  $(a,b)$] and the image of the point $(0,1)$ [call it $(c,d)$]. The matrix is then
$$\begin{bmatrix}
  a & c \\
  b & d \\
\end{bmatrix}$$
Here is one way to find $(a,b)$. The line between $(1,0)$ and $(a,b)$ is perpendicular to the line $y=mx$: what does that say about the slope of the line between $(1,0)$ and $(a,b)$? Also, the midpoint of the line segment between $(1,0)$ and $(a,b)$ is on the line $y=mx$.
Each of those two statements gives you a linear equation in $a$ and $b$. Solve those two simultaneous equations and you have your $a$ and $b$. Do almost the same thing to find $c$ and $d$.
There are other ways to find your matrix, of course, but this way uses no advanced concepts.
