$25 \cdot 21$ In parallelogram $P Q R S$, three vertices are $P(-3,-2), Q(1,-5), R(9,1)$, where $P$ and $R$ are opposite vertices. The sum of the coordinates of $S$ is
(A) 13.
(B) 12.
(C) 11.
(D) 10.
(E) 9.
(14th American High School Mathematics Examination, 1963)
$[$ [Solution] Let $G$ be the intersection point of the diagonals of $\square P Q R S$, then $G$ is the midpoint of $P R$ and $Q S$. By the section formula, we have
$$
\begin{array}{c}
x_{G}=\frac{x_{P}+x_{R}}{2}=\frac{-3+9}{2}=3, \\
y_{G}=\frac{y_{P}+y_{R}}{2}=\frac{-2+1}{2}=-\frac{1}{2} .
\end{array}
$$

Similarly, $3=\frac{x_{S}+x_{Q}}{2}=\frac{x_{S}+1}{2}$,
we get $x_{S}=5$,
and $-\frac{1}{2}=\frac{y_{S}+y_{Q}}{2}=\frac{y_{S}-5}{2}$, we get $y_{S}=4$.
Thus, $x_{S}+y_{S}=9$,
so the answer is $(E)$.