The problem asks which rotation about the origin transforms the point $A(2, -6)$ to the point $A'(-6, -2)$. The possible rotations are: (a) $-180^\circ$, (b) $-90^\circ$, (c) $90^\circ$, and (d) $180^\circ$.

1. Problem Description

The problem asks which rotation about the origin transforms the point $A(2, -6)$ to the point $A'(-6, -2)$. The possible rotations are: (a) $-180^\circ$, (b) $-90^\circ$, (c) $90^\circ$, and (d) $180^\circ$.

2. Solution Steps

A rotation of $\theta$ about the origin transforms a point $(x, y)$ to $(x', y')$, where:

$x' = x \cos \theta - y \sin \theta$

$y' = x \sin \theta + y \cos \theta$

Let's check each option:

(a) $\theta = -180^\circ$: $\cos(-180^\circ) = -1$, $\sin(-180^\circ) = 0$.

$x' = (2)(-1) - (-6)(0) = -2$

$y' = (2)(0) + (-6)(-1) = 6$

This gives $(-2, 6)$, which is not $(-6, -2)$.

(b) $\theta = -90^\circ$: $\cos(-90^\circ) = 0$, $\sin(-90^\circ) = -1$.

$x' = (2)(0) - (-6)(-1) = -6$

$y' = (2)(-1) + (-6)(0) = -2$

This gives $(-6, -2)$, which matches the desired point.

(c) $\theta = 90^\circ$: $\cos(90^\circ) = 0$, $\sin(90^\circ) = 1$.

$x' = (2)(0) - (-6)(1) = 6$

$y' = (2)(1) + (-6)(0) = 2$

This gives $(6, 2)$, which is not $(-6, -2)$.

(d) $\theta = 180^\circ$: $\cos(180^\circ) = -1$, $\sin(180^\circ) = 0$.

$x' = (2)(-1) - (-6)(0) = -2$

$y' = (2)(0) + (-6)(-1) = 6$

This gives $(-2, 6)$, which is not $(-6, -2)$.

Therefore, the rotation that transforms $A(2, -6)$ to $A'(-6, -2)$ is a rotation of $-90^\circ$ about the origin.

3. Final Answer

(b) R (O, -90°)