10. Given point $P$ is on the line $x+2 y-1=0$, point $A$ is on the line $x+2 y+3=0$, the midpoint of $P Q$ is $M\left(x_{0}, y_{0}\right)$, and $y_{0}>x_{0}+2$, then the range of $\frac{y_{0}}{x_{0}}$ is $\qquad$ .
Answer $\left(-\frac{1}{2},-\frac{1}{5}\right)$.
Analysis Note that the two lines are parallel, so the locus of points is a line parallel to the two lines and equidistant from them, with the equation $x+2 y+1=0$, i.e., $M\left(x_{0}, y_{0}\right)$ satisfies $x_{0}+2 y_{0}+1=0$. The points that satisfy the inequality $y_{0}>x_{0}+2$ are in the upper left of the line $y=x+2$. The problem is transformed into finding the range of $\frac{y_{0}}{x_{0}}$ for the point $M\left(x_{0}, y_{0}\right)$ on the ray $x_{0}+2 y_{0}+1=0\left(x_{0}<-\frac{5}{3}\right)$, and the geometric meaning of $\frac{y_{0}}{x_{0}}$ is the slope of the line connecting $M\left(x_{0}, y_{0}\right)$ and the origin, so $k_{O M}=\frac{y_{0}}{x_{0}} \in\left(-\frac{1}{2},-\frac{1}{5}\right)$.