To find the matrix corresponding to a linear transformation from to , it is enough to see what it does to the "basis vectors", <1, 0> and <0, 1>.
so what the linear transformation does to <1, 0> and <0, 1> gives you the two columns of the matrix.
Now, the line, through the origin with slope m, y= mx, contains the points (0, 0) and (1, m) and so is in the direction of the vector <1, m>. To find the "reflection" of <1, 0> in that line, find its projection onto the vector <1, m>, then subtract that from <1, 0> to find its component perpendicular to the line. Subtract twice that vector from <1, 0> (subtracting once puts you on the line, subtracting again, on the other side) to find the reflection in the line.
Alternative 3 :
Some courses of Linear Algebra cover the following and well known theorem:
If is a hyperplane of , then the projection matrix on with the usual inner product is:
As a consequence the simmetry matrix on is
In our case, so,