1. Given $I$ is the incenter of $\triangle A B C$, $A C=2, B C=3$, $A B=4$. If $\overrightarrow{A I}=x \overrightarrow{A B}+y \overrightarrow{A C}$, then $x+y=$ $\qquad$
$-1 . \frac{2}{3}$.
Connect $A I$ and extend it to intersect $B C$ at point $D$. Then $D$ divides the segment $B C$ in the ratio $\lambda=\frac{A B}{A C}=\frac{4}{2}=2$.
Thus, $\overrightarrow{A D}=\frac{1}{3} \overrightarrow{A B}+\frac{2}{3} \overrightarrow{A C}$.
Since $B C=3$, we have $B D=2, D C=1$.
In $\triangle A B D$, $I$ divides the segment $A D$ in the ratio
$$
\lambda^{\prime}=\frac{A B}{B D}=\frac{4}{2}=2 \text {, }
$$

which means $\overrightarrow{A I}=\frac{2}{3} \overrightarrow{A D}=\frac{2}{9} \overrightarrow{A B}+\frac{4}{9} \overrightarrow{A C}$.
Therefore, $x+y=\frac{2}{3}$.