+118608 A midpoint of 2 points is the (average value of the x's, average value of the y's)
T is on the x axis (y=0)
U and V are both on the line y=6,
So R, W and S must be on the line y=12
And P must have a y value of 24.
Mmm
Here is a graphical solution. (It is not the logic solution that you need though.)
It appears that Dragan has beaten me to it. Nice work Dragan. 
But neither of us has provided an answer with explained logic.
Maybe Dragan does have one. Do you Dragan?
+26367 In the diagram, O is lacated at (0,0), U is located at (7,6),
and V is located at (11,6).
Assume that R, S, T, U, V, and W are midpoints.
\(1)\\ \begin{array}{|rcl|rcl|rcl|} \hline R &=& U + \dfrac{TR}{2} & \dfrac{TR}{2}+T &=& U & T &=& \dfrac{Q}{2} \\ && & \dfrac{TR}{2}+\dfrac{Q}{2} &=& U \\ && & \dfrac{TR}{2} &=& U-\dfrac{Q}{2} \\ R &=& U + U-\dfrac{Q}{2} \\ \mathbf{R} &=& \mathbf{2U-\dfrac{Q}{2}} \\ \hline S &=& T+TS & T+\dfrac{TS}{2} &=& V & T &=& \dfrac{Q}{2} \\ & & & \dfrac{Q}{2}+\dfrac{TS}{2} &=& V \quad | \quad *2 \\ & & & Q+TS &=& 2V \\ & & & TS &=& 2V-Q \\ S &=& \dfrac{Q}{2} + 2V-Q \\ \mathbf{S} &=& \mathbf{2V-\dfrac{Q}{2}} \\ \hline \end{array}\)
\(2)\\ \begin{array}{|rcl|rcl|rcl|} \hline && &P &=& S + \dfrac{QP}{2} & QP &=& 2(S-Q) \\ && &P &=& S + (S-Q) \\ && &P &=& 2S-Q & S&=&2V-\dfrac{Q}{2} \\ && &P &=& 4V-Q-Q \\ && &\mathbf{P} &=& \mathbf{4V-2Q} \\ \hline && &P &=& 2R & R&=&2U-\dfrac{Q}{2} \\ && &\mathbf{P} &=& \mathbf{4U-Q} \\ P= 4U-Q &=& 4V-2Q \\ 4U-Q &=& 4V-2Q \\ \mathbf{ Q }&=& \mathbf{4V-4U} \\ \hline \end{array} \\ 3)\\ \begin{array}{|rcl|rcl|rcl|} \hline W&=& \dfrac{1}{2}( R+S )& R&=&2U-\dfrac{Q}{2} & S&=&2V-\dfrac{Q}{2} \\ W&=& \dfrac{1}{2}( 2U-\dfrac{Q}{2}+2V-\dfrac{Q}{2} ) \\ W&=& \dfrac{1}{2}( 2U+2V-Q ) \\ \mathbf{ W }&=& \mathbf{U+V-\dfrac{Q}{2}} \\ \hline \end{array}\)
\(\text{Summary:}\quad Q=4V-4U \\ \begin{array}{|rcll|} \hline T&=& \dfrac{Q}{2} \\ R&=& 2U-\dfrac{Q}{2} \\ S&=& 2V-\dfrac{Q}{2} \\ W&=& U+V-\dfrac{Q}{2} \\ P&=& 4U-Q \\ \hline \end{array}\)
Substitute \(Q\):
\(\begin{array}{|rcll|} \hline T &=& -2U+2V \\ Q &=& -4U+4V \\ R &=& 4U-2V \\ S &=& 2U \\ W &=& 3U-V \\ P &=& 8U-4V \\ \hline \end{array}\)
Substitute \(U(7,6)\) and \(V(11,6)\):
\(\begin{array}{|rcll|} \hline \mathbf{T} &=& -2(7,6)+2(11,6) = (-14+22, -12+12) =\mathbf{(8,0)}\\ \mathbf{Q} &=& -4(7,6)+4(11,6) = (-28+44, -24+24) =\mathbf{(16,0)}\\ \mathbf{R} &=& 4(7,6)-2(11,6) = (28-22, 24-12) =\mathbf{(6,12)}\\ \mathbf{S} &=& 2(7,6) = \mathbf{(14, 12)} \\ \mathbf{W} &=& 3(7,6)-(11,6) = (21-11, 18-6) =\mathbf{(10,12)}\\ \mathbf{P} &=& 8(7,6)-4(11,6) = (56-44, 48-24) =\mathbf{(12,24)}\\ \hline \end{array}\)
