$27 \cdot 9$ Measuring the weights $y$ and heights $x$ of three friends, we obtain the points $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$, $\left(x_{3}, y_{3}\right)$. If $x_{1}<x_{2}<x_{3}$ and $x_{3}-x_{2}=x_{2}-x_{1}$, which of the following numbers is the slope of the line that is “closest” to these three points (where “closest” means that the sum of the squares of the distances from these three points to this line is less than the sum of the squares of the distances from these points to any other line)?
(A) $\frac{y_{3}-y_{1}}{x_{3}-x_{1}}$.
(B) $\frac{\left(y_{2}-y_{1}\right)-\left(y_{3}-y_{2}\right)}{x_{3}-x_{1}}$.
(C) $\frac{2 y_{3}-y_{1}-y_{2}}{2 x_{3}-x_{1}-x_{2}}$.
(D) $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}+\frac{y_{3}-y_{2}}{x_{3}-x_{2}}$.
(E) None of the above.
(39th American High School Mathematics Examination, 1988)
[Solution] Let the required line be $y=a+b x$. According to the problem, we need to minimize
$$
Q(a, b)=\frac{1}{n} \sum_{i=1}^{n}\left(y_{i}-a-b x_{i}\right)^{2}
$$

By completing the square, we can find
$$
a=\bar{y}-\frac{\sigma_{x y}}{\sigma_{x}^{2}} \bar{x}, \quad b=\frac{\sigma_{x y}}{\sigma_{x}^{2}} .
$$

Here, $\quad \bar{x}=\frac{1}{n} \sum_{i=1}^{n} x_{i}, \bar{y}=\frac{1}{n} \sum_{i=1}^{n} y_{i}$,
and $\sigma_{x}^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}, \sigma_{x y}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)\left(y_{i}-\bar{y}\right)$.
Given that $\quad x_{3}-x_{2}=x_{2}-x_{1}$, let $x_{2}=a, \quad x_{2}-x_{1}=h$, then
$$
x_{1}=a-h, \quad x_{3}=a+h .
$$

Thus,
$$
\begin{array}{l}
\bar{x}=\frac{1}{3}[(a-h)+a+(a+h)]=\frac{1}{3} a, \\
\bar{y}=\frac{1}{3}\left(y_{1}+y_{2}+y_{3}\right) .
\end{array}
$$

Also, $\sigma_{x}^{2}=\frac{1}{3}\left(h^{2}+0+h^{2}\right)=\frac{2}{3} h^{2}$,
and $\sigma_{x y}=\frac{1}{3}\left[-h\left(y_{1}-\bar{y}\right)+h\left(y_{3}-\bar{y}\right)\right]=\frac{h}{3}\left(y_{3}-y_{1}\right)$,
so the slope of the required line is
$$
b=\frac{\sigma_{x y}}{\sigma_{x}^{2}}=\frac{\frac{h}{3}\left(y_{3}-y_{1}\right)}{\frac{2}{3} h^{2}}=\frac{y_{3}-y_{1}}{2 h}=\frac{y_{3}-y_{1}}{x_{3}-x_{1}} .
$$

Therefore, the answer is $(A)$.
Note: This problem can also be solved using the method of "least squares polynomial fitting".