$27 \cdot 1$ Given two points $P(-1,-2)$ and $Q(4,2)$ on the $xy$ plane, take point $R(1, m)$, such that $PR + RQ$ is minimized, then $m$ equals
(A) $-\frac{3}{5}$.
(B) $-\frac{2}{5}$.
(C) $-\frac{1}{5}$.
(D) $\frac{1}{5}$.
(E) $-\frac{1}{5}$ or $\frac{1}{5}$.
(20th American High School Mathematics Examination, 1969)
[Solution] To minimize $P R+R Q$, it is only necessary for $P, R, Q$ to be collinear.
Thus,
$$
\frac{m-(-2)}{1-(-1)}=\frac{2-(-2)}{4-(-1)},
$$
we get
$$
m=-\frac{2}{5} \text{. }
$$

Therefore, the answer is $(B)$.