To determine what happens to a point, P, when we rotate it about a point, C, that is not at the origin of the Cartesian plane, O; we can first translate C to O, then rotate about O, then undo the translation. See how this works in the example below.

Now, let's try using conjugation to determine what happens when we reflect over a line that does not pass through the origin.

[math]P\left(x,y\right)\longrightarrow P'\left(x,y-3\right)[/math]

To rotate, we use a rotation around the origin by [math]-\theta[/math] where [math]\theta=tan^{-1}\left(\sqrt{3}\right)=60^\circ[/math].

[math]P''\left(0.5x+\frac{\sqrt{3}}{2}\left(y-3\right),-\frac{\sqrt{3}}{2}x+0.5\left(y-3\right)\right)\longrightarrow P'''\left(0.5x+\frac{\sqrt{3}}{2}\left(y-3\right),\frac{\sqrt{3}}{2}x-0.5\left(y-3\right)\right)[/math]

[math]P''''\left(\frac{1}{4}x+\frac{\sqrt{3}}{4}\left(y-3\right)-\frac{3}{4}x+\frac{\sqrt{3}}{4}\left(y-3\right),\frac{\sqrt{3}}{4}x+\frac{3}{4}\left(y-3\right)+\frac{\sqrt{3}}{4}x-\frac{1}{4}\left(y-3\right)\right)\longrightarrow P'''''\left(\frac{1}{4}x+\frac{\sqrt{3}}{4}\left(y-3\right)-\frac{3}{4}x+\frac{\sqrt{3}}{4}\left(y-3\right),\frac{\sqrt{3}}{4}x+\frac{3}{4}\left(y-3\right)+\frac{\sqrt{3}}{4}x-\frac{1}{4}\left(y-3\right)+3\right)[/math]