2. If the line $y=k x+1$ intersects the circle $x^{2}+y^{2}+k x+$ $m y-4=0$ at points $P$ and $Q$, and points $P$ and $Q$ are symmetric with respect to the line $x+y=0$, then the area of the plane region represented by the system of inequalities
$$
\left\{\begin{array}{l}
k x-y+1 \geqslant 0, \\
k x-m y \leqslant 0, \\
y \geqslant 0
\end{array}\right.
$$

is ( ).
(A) 2
(B) 1
(C) $\frac{1}{2}$
(D) $\frac{1}{4}$
2.D.

From the fact that points $P$ and $Q$ are symmetric with respect to the line $x+y=0$, we know that the line $y=kx+1$ is perpendicular to $x+y=0$, and the center $\left(-\frac{k}{2},-\frac{m}{2}\right)$ of the circle lies on the line $x+y=0$, which gives $k=1$, and $\left(-\frac{k}{2}\right)+\left(-\frac{m}{2}\right)=0$. Solving these equations simultaneously, we get $k=1, m=-1$. Therefore, the system of inequalities $\left\{\begin{array}{l}x-y+1 \geqslant 0, \\ x+y \leqslant 0, \\ y \geqslant 0\end{array}\right.$ represents the plane region of the isosceles right triangle $\triangle ABO$ in Figure 3, with an area of $S=\frac{1}{4}$.