9. Given point $P(2,1)$, draw a line $l$ through $P$ that intersects the positive $x$-axis and $y$-axis at points $A$ and $B$ respectively. The equation of the line $l$ that minimizes the perimeter of $\triangle A O B$ (where $O$ is the origin) is $\qquad$

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$$
9.3 x+4 y-10=0 \text {. }
$$

Let $\angle B A O=\theta, t=$ $\tan \frac{\theta}{2}$. Denote the perimeter of $\triangle A B O$ as $s$, then
$$
\begin{aligned}
s= & (2+\cot \theta)+ \\
& (1+2 \tan \theta)+ \\
& \left(\frac{1}{\sin \theta}+\frac{2}{\cos \theta}\right) \\
= & 6+\frac{1-t}{t}+\frac{4 t}{1-t} \geqslant 10 .
\end{aligned}
$$

Equality holds if and only if $\frac{1-t}{t}=\frac{4 t}{1-t}$, i.e., $\tan \frac{\theta}{2}=\frac{1}{3}$.
Then the slope of the line $l$ is $k=-\frac{3}{4}$.
Therefore, the equation of the line $l$ is $3 x+4 y-10=0$.