Example 3. Point $\mathrm{P}$ is a moving point on a fixed line segment $\mathrm{AB}$, where $\mathrm{AB}=\mathrm{a}$. On the same side of $\mathrm{AB}$, equilateral triangles $\mathrm{APM}$ and $\mathrm{BPN}$ are constructed with $\mathrm{AP}$ and $\mathrm{PB}$ as sides, respectively. Find the locus of the midpoint $\mathrm{Q}$ of line segment $\mathrm{MN}$.
Analysis: As shown in Figure 3, with point $\mathrm{A}$ as the origin and line $\mathrm{AB}$ as the $\mathrm{x}$-axis, we establish a coordinate system. Clearly, the position of $\mathrm{Q}$ depends on the position of the moving point $\mathrm{P}$ on $\mathrm{AB}$, i.e., the distance from point $\mathrm{P}$ to the origin. Therefore, if we set the coordinates of point $\mathrm{P}$ as $(t, 0)$, with $t$ as the parameter, then from the given information, we can find the coordinates of point $\mathrm{M}$ to be: $\left(\frac{t}{2}, \frac{\sqrt{3}}{2} t\right)$, and the coordinates of point $\mathrm{N}$ to be $\left(t+\frac{1}{2}(a-\right.$ t), $\frac{\sqrt{3}}{2}(a-t)$ ). Thus, the coordinates of the midpoint $Q(x, y)$ of $M N$ are: $x=\frac{t}{2}+\frac{a}{4}, y=\frac{\sqrt{3}}{4} \mathrm{a}$, which is the parametric equation of the trajectory of point $Q$. Since $y$ is already a constant (not containing the parameter), it represents a line parallel to the $x$-axis. Also, from $0<t<a$, we know $\frac{a}{4}<\frac{t}{2}+\frac{a}{4} < \frac{8}{4} a$. Therefore, the trajectory equation of $\mathrm{Q}$ is $\mathrm{y}=\frac{\sqrt{3}}{4} \mathrm{a}, \quad\left(\frac{\mathrm{a}}{4}<\mathrm{x}<\right.$ $\left.\frac{3}{4} \mathrm{a}\right)$, indicating that the required trajectory is a line segment with endpoints $\left(\frac{\mathrm{a}}{4}\right.$, $\left.\frac{\sqrt{3}}{4} \mathrm{a}\right)$ and $\left(\frac{3}{4} \mathrm{a}, \frac{\sqrt{3}}{4} \mathrm{a}\right)$ (excluding the endpoints).