5.9 As shown in the figure, the graphs of the direct proportion functions $y=x$ and $y=a x(a>0)$ intersect with the graph of the inverse proportion function $y=\frac{k}{x}(k>0)$ at points $A$ and $C$, respectively. If the areas of the right triangles $\triangle A O B$ and $\triangle C O D$ are $S_{1}$ and $S_{2}$, respectively, then the relationship between $S_{1}$ and $S_{2}$ is
(A) $S_{1}>S_{2}$.
(B) $S_{1}=S_{2}$.
(C) $S_{1}<S_{2}$.
(D) Uncertain.
[Solution] Let the coordinates of point $A$ be $\left(x_{1}, y_{1}\right)$, and the coordinates of point $C$ be $\left(x_{2}, y_{2}\right)$.
Then $x_{1} y_{1}=k, x_{2} y_{2}=k$.
Also, $S_{1}=\frac{1}{2} x_{1} y_{1}, S_{2}=\frac{1}{2} x_{2} y_{2}$, so $S_{1}=S_{2}$. Therefore, the answer is $(B)$.