2. In the Cartesian coordinate system, the area of the plane region represented by the system of inequalities $\left\{\begin{array}{l}\sqrt{3} x-y \leqslant 0, \\ x-\sqrt{3} y+2 \geqslant 0, \\ y \geqslant 0\end{array}\right.$ is
A. $\frac{\sqrt{3}}{2}$
B. $\sqrt{3}$
C. 2
D. $2 \sqrt{3}$
The plane region has three vertices with coordinates $A(1, \sqrt{3}), B(0,0), C(-2,0)$, and its area is $S=\sqrt{3}$, so the answer is $B$.

