Rotation in R2, around a line?

[DDrew]

Homework Statement

Find the points fixed by f, and show it is a line. We know that f is an isometry.

0 $$\leq$$ $$\theta$$ < 2$$\Pi$$
f: R$$^{2}$$ $$\rightarrow$$ R$$^{2}$$
f(x) = Ax

A = | cos $$\theta$$ sin $$\theta$$ |
...| sin $$\theta$$ -cos $$\theta$$|

Homework Equations

fix(f) = {x | f(x) = x}

The Attempt at a Solution

fix(f) = | (x1)(cos $$\theta$$) + (x2)(sin $$\theta$$) = x1 |
...| (x1)(sin $$\theta$$) + (x2)(cos $$\theta$$) = x2 |

The thing that confuses me the most is rotating about a line in R$$^{2}$$. I was under the impression you could only rotate around a point? Is there something I'm missing? I'm not looking for the answer so much as I'm looking to be pointed in the right direction.

I apologize for my messy matrix formatting.

 

[lippyka]

A:Let's look at the first equation:$$(x_1)\cos \theta + (x_2)\sin \theta = x_1.$$We can solve it for $x_2$: $$x_2 = \frac{x_1- (x_1)\cos \theta}{\sin \theta}.$$So, in order for this equation to be satisfied for any value of $x_1$, the denominator must be zero, i.e. $\sin \theta=0$. But $$\sin \theta=0 \iff \theta = n\pi, n\in \mathbb Z.$$So, our isometry fixes the points on the line $$x_2 = n\pi.$$