2. Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$ be any two points on the graph of the function $f(x)=\frac{1}{2}+\log _{2} \frac{x}{1-x}$, and $\overrightarrow{O M}=\frac{1}{2}(\overrightarrow{O A}+\overrightarrow{O B})$. Given that the x-coordinate of point $M$ is $\frac{1}{2}$, then the y-coordinate of point $M$ is $\qquad$.
2. $\frac{1}{2}$ Analysis: Since $\overrightarrow{O M}=\frac{1}{2}(\overrightarrow{O A}+\overrightarrow{O B})$, $M$ is the midpoint of $A B$. Let $M(x, y)$. From $\frac{1}{2}\left(x_{1}+x_{2}\right)=x=\frac{1}{2}$, we get $x_{1}+x_{2}=1$. And $y_{1}+y_{2}=\frac{1}{2}+\log _{2} \frac{x_{1}}{1-x_{1}}+\frac{1}{2}+\log _{2} \frac{x_{2}}{1-x_{2}}=1+\log _{2}\left(\frac{x_{1}}{1-x_{1}} \cdot \frac{x_{2}}{1-x_{2}}\right)=1+$ $\log _{2} \frac{x_{1} \cdot x_{2}}{x_{1} \cdot x_{2}}=1$, so $y=\frac{y_{1}+y_{2}}{2}=\frac{1}{2}$.