$25 \cdot 26$ A line passes through the point $(-a, 0)$ and divides the second quadrant into a triangular region with area $T$. The equation of this line is
(A) $2 T x+a^{2} y+2 a T=0$.
(B) $2 T x-a^{2} y+2 a T=0$.
(C) $2 T x+a^{2} y-2 a T=0$.
(D) $2 T x-a^{2} y-2 a T=0$.
(E) None of the above.
(15th American High School Mathematics Examination, 1964)
[Solution] Clearly, the intersection point of the line with the $x$-axis is $(-a, 0)$, and let its intersection point with the $y$-axis be $(0, b)$. Given that 
$S_{\triangle O A B}=\frac{1}{2} a b=T$, then $b=\frac{2 T}{a}$.
Thus, the slope of the line is $\frac{b}{a}=\frac{2 T}{a^{2}}$, and the equation of the line is
$$
y=\frac{2 T}{a^{2}} x+\frac{2 T}{a},
$$

or $2 T x-a^{2} y+2 a T=0$.
Therefore, the answer is $(B)$.