9. A. Given two inverse proportion functions $y=\frac{k}{x}$ and $y=\frac{1}{x}$, their graphs in the first quadrant are shown in Figure 4. Point $P$ is on the graph of $y=\frac{k}{x}$, $P C \perp x$-axis, with the foot of the perpendicular at $C$, and intersects the graph of $y=\frac{1}{x}$ at point $A$; $P D \perp y$-axis, with the foot of the perpendicular at $D$, and intersects the graph of $y=\frac{1}{x}$ at point $B$. When point $P$ moves on the graph of $y=\frac{k}{x}$, the following conclusions are:
(1) The areas of $\triangle O D B$ and $\triangle O C A$ are equal;
(2) The area of quadrilateral $O A P B$ remains constant;
(3) $P A$ and $P B$ are equal;
(4) When $A$ is the midpoint of $P C$, $B$ is the midpoint of $P D$. Among these, the correct conclusions are $\qquad$ (fill in the sequence numbers of all correct conclusions).

Figure 4
9. A. (1)(2)(4).

Let $P\left(a, \frac{k}{a}\right)$ be any point on the graph of the function $y=\frac{k}{x}$ in the first quadrant. Then $A\left(a, \frac{1}{a}\right), B\left(\frac{a}{k}, \frac{k}{a}\right)$.
Thus, $S_{\text {quadrilateral } O A P B}=S_{\text {rectangle } O C P D}-\left(S_{\triangle A O C}+S_{\triangle B O D}\right)$ $=a \cdot \frac{k}{a}-1=k-1$.
If $A$ is the midpoint of $P C$, then
$$
\frac{k}{2 a}=\frac{1}{a} \Rightarrow k=2 \text {. }
$$

Therefore, $B\left(\frac{a}{2}, \frac{2}{a}\right)$ is the midpoint of $P D$.
Note that, $P A=\frac{k}{a}-\frac{1}{a}=\frac{k-1}{a}$,
$$
P B=a-\frac{a}{k}=\frac{a(k-1)}{k} .
$$

$P A$ and $P B$ are equal only when $a=\sqrt{k}$.
In summary, conclusions (1)(2)(4) are correct.