10. (20 points) Given an ellipse centered at the origin, with foci on the $x$-axis, the length of the major axis is twice the length of the minor axis, and it passes through the point $M(2,1)$. A line $l$ parallel to $OM$ has a $y$-intercept of $m (m<0)$, and intersects the ellipse at two distinct points $A$ and $B$. Find the $x$-coordinate of the incenter $I$ of $\triangle ABM$.

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10. Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$.
Then $\left\{\begin{array}{l}a=2 b, \\ \frac{4}{a^{2}}+\frac{1}{b^{2}}=1\end{array} \Rightarrow\left\{\begin{array}{l}a^{2}=8, \\ b^{2}=2 \text {. }\end{array}\right.\right.$
Therefore, the equation of the ellipse is $\frac{x^{2}}{8}+\frac{y^{2}}{2}=1$.
As shown in Figure 1, because

the line $l$ is parallel
to $O M$, and the y-intercept
is $m$, and $k_{O S}=\frac{1}{2}$,
so, the equation of the line $l$ is $y=\frac{1}{2} x+m$.
From $\left\{\begin{array}{l}y=\frac{1}{2} x+m, \\ \frac{x^{2}}{8}+\frac{y^{2}}{2}=1\end{array} \Rightarrow x^{2}+2 m x+2 m^{2}-4=0\right.$.
Let $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right)$. Then
$x_{1}+x_{2}=-2 m, x_{1} x_{2}=2 m^{2}-4$.
Let the slopes of the lines $M A 、 M B$ be $k_{1} 、 k_{2}$. Then
$k_{1}=\frac{y_{1}-1}{x_{1}-2}, k_{2}=\frac{y_{2}-1}{x_{2}-2}$.
Thus, $k_{1}+k_{2}=\frac{y_{1}-1}{x_{1}-2}+\frac{y_{2}-1}{x_{2}-2}$
$=\frac{\left(y_{1}-1\right)\left(x_{2}-2\right)+\left(y_{2}-1\right)\left(x_{1}-2\right)}{\left(x_{1}-2\right)\left(x_{2}-2\right)}$.
The numerator of the above expression
$$
\begin{array}{l}
=\left(\frac{1}{2} x_{1}+m-1\right)\left(x_{2}-2\right)+\left(\frac{1}{2} x_{2}+m-1\right)\left(x_{1}-2\right) \\
=x_{1} x_{2}+(m-2)\left(x_{1}+x_{2}\right)-4(m-1) \\
=2 m^{2}-4+(m-2)(-2 m)-4(m-1) \\
=0 .
\end{array}
$$

Therefore, $k_{1}+k_{2}=0$.
Hence, the angle bisector $M I$ of $\triangle A B M$ is perpendicular to the x-axis. Therefore, the x-coordinate of the incenter $I$ is equal to the x-coordinate of point $M$, which is 2.