imc

Due to rotations preserving an equal distance, we can bash the answer with the distance formula. $D(A,P)=D(A^{\prime },P)$, and $D(B,P)=D(B^{\prime },P)$. Thus we will square our equations to yield: $(1-r{)}^2+(2-s{)}^2=(3-r{)}^2+(1-s{)}^2$, and $(3-r{)}^2+(3-s{)}^2=(4-r{)}^2+(3-s{)}^2$. Canceling $(3-s{)}^2$ from the second equation makes it clear that $r$ equals $3.5$. Substituting will yield $$\begin{array}{rl}(2.5{)}^2+(2-s{)}^2& =(-0.5{)}^2+(1-s{)}^2\\ 6.25+4-4s+s^2& =0.25+1-2s+s^2\\ 2s& =9\\ s& =4.5\end{array}$$. Now $|r-s|=|3.5-4.5|=\boxed{1}$.