Show that $x$ and $Q x$ are equidistant $$Q= \begin{bmatrix} \cos x & -\sin x\\ \sin x & \cos x\end{bmatrix}$$
Given x belongs to $\mathbb{R^2}$, show $Qx$ and $x$ are equidistant.
I've tried dot producting $Qx$ and seeing whether they are equal. I just can't seem to get it.

Of course, @thanasissdr answer is the well educated one, but arguing more naïvely:
As written in my comment:
$$d((a,b),(0,0)=\sqrt{(a-0)^2+(b-0)^2}=\sqrt{a^2+b^2}=\|(a,b)\|$$
Thus we compare the norms.
We have:
$$\|(a,b)\|=\sqrt{a^2+b^2}$$
and
\begin{align*}\|Q(x,y)\|& =\|(\cos(x)a+\sin(x)b,-\sin(x)a+\cos(x)b)\|\\ & =\sqrt{(\cos(x)a+\sin(x)b)^2+(-\sin(x)a+\cos(x)b)^2}\\ & =\sqrt{\cos^2(x)a^2+\sin^2(x)b^2+2\sin(x)\cos(x)ab\cos^2(x)b^2+\sin^2(x)a^2\color{red}{-}2\sin(x)\cos(x)ab}\\ & =\sqrt{[\sin^2(x)+ \cos^2(x)]a^2+[\sin^2(x)b^2+\cos^2(x)]b^2}\\ & =\sqrt{a^2+b^2}\end{align*}
as claimed, where we have used the classic identity $\sin^2(x)+\cos^2(x)=1$.
Note: Any matrix $Q$ which fulfills this property is called orthogonal matrix. This is equivalent to the above cited criterion $Q^\top Q=1$.
