5. As shown in Figure 2, in the Cartesian coordinate system $x O y$, the vertices of the isosceles trapezoid $A B C D$ have coordinates $A(1,1)$, $B(2,-1)$, $C(-2,-1)$, and $D(-1,1)$. A point $P(0,2)$ on the $y$-axis is rotated $180^{\circ}$ about point $A$ to get point $P_{1}$, $P_{1}$ is rotated $180^{\circ}$ about point $B$ to get point $P_{2}$, $P_{2}$ is rotated $180^{\circ}$ about point $C$ to get point $P_{3}$, and $P_{3}$ is rotated $180^{\circ}$ about point $D$ to get point $P_{4}$, and so on. Repeating this operation, points $P_{1}, P_{2}, \cdots, P_{2010}$ are obtained. What are the coordinates of point $P_{2010}$? ( . ).
(A) $(2010,2)$
(B) $(2010,-2)$
(C) $(2012,-2)$
(D) $(0,2)$
5. B.

From the given, we have points $P_{1}(2,0)$ and $P_{2}(2,-2)$. Let $P_{2}\left(a_{2}, b_{2}\right)$, where $a_{2}=2, b_{2}=-2$. According to the symmetry relationship, we sequentially obtain
$$
\begin{array}{l}
P_{3}\left(-4-a_{2},-2-b_{2}\right), \\
P_{4}\left(2+a_{2}, 4+b_{2}\right), \\
P_{5}\left(-a_{2},-2-b_{2}\right), P_{6}\left(4+a_{2}, b_{2}\right) . \\
\text { Let } P_{6}\left(a_{6}, b_{2}\right) .
\end{array}
$$

Similarly, point $P_{10}\left(4+a_{6}, b_{2}\right)$, which is
$$
\begin{array}{l}
P_{10}\left(4 \times 2+a_{2}, b_{2}\right) \text {. } \\
\text { Since } 2010=4 \times 502+2 \text {, then } \\
P_{2010}(2010,-2) \text {. }
\end{array}
$$