9. Let the line $l: y=x+b\left(0<b<\frac{1}{2}\right)$ intersect the parabola $y^{2}=2 x$ at points $A$ and $B$, and let $O$ be the origin. Then, when the area of $\triangle A O B$ is maximized, the equation of the line $l$ is $\qquad$
9. $y=x+\frac{1}{3}$.
$|A B|=2 \sqrt{2} \cdot \sqrt{1-2 b}$, the distance from $0$ to $A B$ is $d=\frac{b}{\sqrt{2}}$, so $S_{\triangle \triangle B B}=b \sqrt{1-2 b}$.
When $b=\frac{1}{3}$, the maximum value is $\frac{\sqrt{3}}{9}$.