Square $PQRS$ lies in the first quadrant. Points $(6,0), (10,0), (14,0),$ and $(26,0)$ lie on lines $SP, RQ, PQ,$ and $SR$, respectively. What is the sum of the coordinates of the center of the square $PQRS$?

1. As in the problem, $SP: y=m(x-6)$ and $RQ: y=m(x-10)$ are parallel, and $PQ: x+my=14$ and $SR: x+my=26$ are perpendicular to them.
2. The distances between the parallel pairs are $\frac{4|m|}{\sqrt{1+m^2}}$ and $\frac{12}{\sqrt{1+m^2}}$; they are equal, so $m=3$.
3. The center lies on the midlines $y=3(x-8)$ and $x+3y=20$, so it is $(\frac{46}5, \frac{18}5)$.
4. $\frac{46}5+ \frac{18}5=\frac{64}5=\boxed{12.8}$.