Question

what is the coordinate for the image of point h(2, -6) under a 90° clockwise rotation about the origin?
○ h(-6, -2)
○ h(6, -2)
○ h(-6, 2)
○ h(6, 2)

Explanation:

Step1: Recall 90° clockwise rotation rule

For a point $(x,y)$, 90° clockwise rotation about the origin transforms it to $(y, -x)$.

Step2: Substitute $x=2, y=-6$

Substitute into the rule: $(y, -x) = (-6, -2)$? No, correct substitution: $(y, -x) = (-6, -2)$ is wrong, correct: $(y, -x) = (-6, - (2))$ → no, wait: 90° clockwise rotation formula is $(x,y) \to (y, -x)$. So $x=2$, $y=-6$: new $x$ is $y=-6$, new $y$ is $-x=-2$? No, wait, correct standard 90° clockwise rotation about origin is $(x,y) \mapsto (y, -x)$. So plugging in:
New $x$-coordinate: $y = -6$
New $y$-coordinate: $-x = -2$
Wait no, wait another way: 90° clockwise is equivalent to 270° counterclockwise, formula is $(x,y) \to (y, -x)$. So yes, $(2,-6) \to (-6, -2)$? No, wait no, let's verify:
Wait no, I mixed up. Correct 90° clockwise rotation about origin: $(x, y) \to (y, -x)$. So for $(2, -6)$:
New $x$: $y = -6$
New $y$: $-x = -2$
Wait but let's check with coordinate plane: point (2,-6) is in 4th quadrant. Rotating 90° clockwise would move it to 3rd quadrant, which has negative x and negative y, which matches $(-6,-2)$? Wait no, wait no, wait 90° clockwise from 4th quadrant: if you face origin, rotate 90° clockwise, (2,-6) would go to (-6,-2)? Wait no, let's use rotation matrix. The rotation matrix for 90° clockwise is

$$\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$$

? No, wait rotation matrix for $\theta$ counterclockwise is

$$\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}$$

. 90° clockwise is $-90°$ counterclockwise, so $\theta=-90°$. $\cos(-90°)=0$, $\sin(-90°)=-1$. So matrix is

$$\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$$

? Wait no:

$$\begin{pmatrix}\cos(-90°) & -\sin(-90°) \\ \sin(-90°) & \cos(-90°)\end{pmatrix} = \begin{pmatrix}0 & -(-1) \\ -1 & 0\end{pmatrix} = \begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$$

Multiply by vector

$$\begin{pmatrix}2 \\ -6\end{pmatrix}$$

:
$0*2 + 1*(-6) = -6$
$-1*2 + 0*(-6) = -2$
Yes, so the new point is $(-6, -2)$.

Answer:

H'(-6, -2)