jmc

When we rotate images $90^{\circ }$ the coordinates switch places, and the signs are adjusted based on whether or not an axis was crossed. In this case, rotating point $A$ $90^{\circ }$ will bring it across the $y$-axis into Quadrant I, which means both the $x$ and $y$ will be positive. The original point $A$ was at $(-4,1)$ so the final image will be at $(1,4)$. We also could solve this problem by seeing that the slope of the segment from the origin to $A$ is $-1/4$. If $A$ is moving to a location that is a $90^{\circ }$ rotation about the origin, it will move to a point on the segment perpendicular to the one that currently connects it to the origin. This will be the segment that has a slope of 4/1 or $-4/-1$ from the origin which puts us at $(1,4)$ or $(-1,-4)$. The point $\boxed{(1,4)}$ is in the clockwise direction we need.