= Solution
Let $v=T(0)$ and $B(x)=T(x)-v$. The <Euclidean isometry> $B$ fixes the origin and preserves norms. The <polarization identity> therefore gives $\langle Bx,By\rangle=\langle x,y\rangle$. In particular, $B e_1,B e_2$ are an <orthonormal basis>. The coefficients of $Bx$ in this basis are $\langle Bx,B e_j\rangle=x_j$, so $Bx=x_1B e_1+x_2B e_2$. Thus $B$ is a <linear map> represented by an <orthogonal matrix>, and \b[$T(x)=Bx+v$.] Conversely, every such map preserves distances because $\|B(x-y)\|=\|x-y\|$.
When $\det B=-1$, the real <eigenvalues> of the two-dimensional <orthogonal matrix> are $1,-1$. Choose corresponding unit <eigenvectors> $e,n$, and write $v=ae+bn$. In these coordinates $T(s,t)=(s+a,-t+b)$. This is reflection in the line $t=b/2$, followed by translation $ae$ along that line. \b[It is a <Euclidean reflection> if $a=0$, and a <glide reflection> if $a\ne0$.]
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