Given points $P,Q,P',Q'\in\mathbb{E}^2$ with $d(P,Q)=d(P',Q')$. Show there exists exactly two motions (isometries) $T_{1},T_{2}$ that maps $P\to P'$ and $Q\to Q'$
How exactly would I attempt proving this? I know that I can choose an appropriate coordinate system so that $P=(0,0)$ and $Q=(1,0)$ but how would I show that there are exactly two? Would it be that you split the two mappings where one is a direct mapping and the other is indirect? Or could you just construct the two mappings by composition (you take $P\to P'$ first and then take $Q\to Q'$ and then the other way around - or would those mappings not be unique?)
Thanks!
Observe that any isometry $T$ sending the pair $(P,Q)$ to $(P',Q')$ also maps the line between $l(P,Q)$ in the line $l(P',Q')$ ($l(A,B)$ being the unique line between the two points $A$ and $B$). Then, if $X\not\in l(P,Q)$, there is only another point $X'$ such that $d(P,X)=d(P,X')$ and $d(Q,X)=d(P,X')$ (namely the reflection with respect to the line). This means that every point has at most two possible images. Once such an image is fixed for one point not on $l(P,Q)$, as you were saying you can assume to have a complete frame of reference. Hence there are at most two isometries.