$27 \cdot 2 A, B$ and $C$ have coordinates $(5,5),(2,1)$ and $(0, k)$, respectively. To make $AC + BC$ as small as possible, the value of $k$ is
(A) 3.
(B) $4 \frac{1}{2}$.
(C) $3 \frac{6}{7}$.
(D) $4 \frac{5}{6}$.
(E) $2 \frac{1}{7}$.
(4th American High School Mathematics Examination, 1953)
[Solution] To minimize $AC + BC$, we first need to find the symmetric point $B'$ of $B$ with respect to the $y$-axis, then connect $A B'$ intersecting the $y$-axis at $C$. Thus, $C$ is the point we are looking for.
Obviously, the equation of $A B'$ is
$$
\frac{y-5}{x-5}=\frac{5-1}{5-(-2)}=\frac{4}{7} \text {, }
$$

Let $x=0$, we get $y=2 \frac{1}{7}$.
Thus, $\square$
$$
k=2 \frac{1}{7} \text {. }
$$

Therefore, the answer is $(E)$.