Which of the following rigid transformations (isometries) maps the line segment $\overline{AB}$ onto the line segment $\overline{A'B'}$ so that the image of $A(-2, 1)$ is $A'(2, -1)$ and the image of $B(-1, 4)$ is $B'(1, -4)$?
$\textbf{(A) }$ reflection in the $y$-axis
$\textbf{(B) }$ counterclockwise rotation around the origin by $90^{\circ}$
$\textbf{(C) }$ translation by 3 units to the right and 5 units down
$\textbf{(D) }$ reflection in the $x$-axis
$\textbf{(E) }$ clockwise rotation about the origin by $180^{\circ}$
We can simply graph the points, or use coordinate geometry, to realize that both $A'$ and $B'$ are, respectively, obtained by rotating $A$ and $B$ by $180^{\circ}$ about the origin. Hence the rotation by $180^{\circ}$ about the origin maps the line segment $\overline{AB}$ to the line segment $\overline{A'B'}$, so the answer is $\boxed{(\text{E})}$.
~Dodgers66