Given two line segments $[P,Q]$ and $[P',Q']$ both in 2D Euclidean frames, with both having the same length, i.e. $d(P,Q) = d(P',Q') \gt 0$ , show that there exists exactly two motions such that $T(P) = P'$ and $T(Q) = Q'$ (where T is the motion).
Now, I know that the $d$ operator only produces positive values. The first motion exists because $d(T(P),T(Q)) = d(P,Q) = d(P',Q')$ would the second simply be $-T$?
Existence: translate by $\vec{PP'}$. Then rotate around $P'$ by $\angle QP'Q'$. This gives us one movement $T_1$. Now reflect at line $P'Q'$ afterwards. This gives us a movement $T_2$. (Why is $T_1\ne T_2$?)
Uniquemess: Assume $T_3$ is a third such motion. Then $T_3^{-1}T_1$ and $T_3^{-1}T_2$ and the identity are three motions with $P$ and $Q$ as fixed points. There are only two such motions (why?), hence two of the three are equal. The only possibility os that one of $T_3^{-1}T_1$, $T_3^{-1}T_2$ equald the identity, hence $T_3$ equals one of $T_1,T_2$.