$25 \cdot 28$ On the $xy$-coordinate plane, a line perpendicular to the $x$-axis divides a triangle with vertices at $(0,0)$, $(1,1)$, $(9,1)$ into two regions of equal area. The equation of this line is $x$ equals
(A) 2.5.
(B) 3.0.
(C) 3.5.
(D) 4.0.
(E) 4.5.
(33rd American High School Mathematics Examination, 1982)
[Solution] As shown in the figure, let's construct $\triangle ABC$ on the coordinate plane, and the line $x=a$ divides $\triangle ABC$ into two regions of equal area.

Given $S_{\triangle ABC}=\frac{1}{2} \cdot 1 \cdot 8=4$, the areas of the two equal regions are both 2.

Since the area of the left part of $\triangle ABC$ divided by the line through point $A$ and perpendicular to the $x$-axis is less than the area of $\triangle ABF = \frac{1}{2}$, the line $x=a$ must be to the right of point $A$. The equation of line $BC$ is $y=\frac{x}{9}$, so the coordinates of point $E$ are $\left(a, \frac{a}{9}\right)$.
Thus, $S_{\triangle DEC}=2=\frac{1}{2}\left(1-\frac{a}{9}\right) \cdot(9-a)$, which simplifies to $(9-a)^{2}=36$. Therefore, $9-a= \pm 6$, yielding $a=15$ or 3.
Since this line must intersect $\triangle ABC$, we take $x=3$. Hence, the answer is $(B)$.