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

Step2: Apply the rule to point \(H(2, -6)\)

Here, \(x = 2\) and \(y = -6\). Substituting into the rule, we get the new \(x\)-coordinate as \(y=-6\) and the new \(y\)-coordinate as \(-x = -2\)? Wait, no, wait. Wait, the correct rule for 90° clockwise rotation is \((x,y)\to(y, -x)\)? Wait, no, let's correct. The standard 90° clockwise rotation about the origin: the transformation is \((x, y)\) becomes \((y, -x)\)? Wait, no, actually, let's derive it. A 90° clockwise rotation is equivalent to a 270° counterclockwise rotation. The rotation matrix for 90° clockwise is \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\). So if we have a point \((x,y)\), multiplying by the matrix: \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}y\\-x\end{pmatrix}\). Wait, but let's test with a point. Let's take \((1,0)\), 90° clockwise should be \((0, -1)\). Using the matrix: \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}0\\-1\end{pmatrix}\), which is correct. Another point: \((0,1)\), 90° clockwise should be \((1, 0)\). Matrix: \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}\), correct. So the rule is \((x, y)\to(y, -x)\). Wait, but our point is \(H(2, -6)\). So \(x = 2\), \(y = -6\). Applying the rule: new \(x = y = -6\)? No, wait, no, I think I messed up. Wait, no, the rotation matrix for 90° clockwise is actually \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\)? Wait, no, let's check the standard rotation matrices. The rotation matrix for \(\theta\) degrees counterclockwise is \(\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\). For 90° counterclockwise, \(\theta = 90°\), \(\cos90° = 0\), \(\sin90° = 1\), so matrix is \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\). For 90° clockwise, that's \(\theta = -90°\) or \(270°\) counterclockwise. So \(\cos(-90°)=0\), \(\sin(-90°)= -1\), so the matrix is \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\). So applying to \((x,y)\): \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}y\\-x\end{pmatrix}\). Wait, but let's take point \((2, -6)\). So \(x = 2\), \(y = -6\). Then the new point is \((y, -x)=(-6, -2)\)? No, that's not matching the options. Wait, maybe I got the rule wrong. Wait, another way: 90° clockwise rotation about the origin: the coordinates \((x, y)\) become \((y, -x)\)? Wait, no, let's take a simple point, say \((1, 0)\). 90° clockwise should be \((0, -1)\). Using \((y, -x)\): \(y = 0\), \(-x = -1\), so \((0, -1)\), correct. Another point: \((0, 1)\). 90° clockwise should be \((1, 0)\). Using \((y, -x)\): \(y = 1\), \(-x = 0\), so \((1, 0)\), correct. Another point: \((2, 3)\). 90° clockwise should be \((3, -2)\). Let's check with the matrix: \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}2\\3\end{pmatrix}=\begin{pmatrix}3\\-2\end{pmatrix}\), correct. So the rule is \((x, y)\to(y, -x)\). Wait, but our point is \(H(2, -6)\). So \(x = 2\), \(y = -6\). Applying the rule: new \(x = y = -6\), new \(y = -x = -2\)? Wait, no, that would be \((-6, -2)\), but that's not one of the options? Wait, the options are:

- \(H'(6, 2)\)

- \(H'(-6, -2)\)

- \(H'(6, -2)\)

- \(H'(-6, 2)\)

Wait, maybe I made a mistake in the rule. Wait, maybe the 90° clockwise rotation rule is \((x, y)\to(y, -x)\)? Wait, no, let's check again. Wait, maybe the correct rule is 90° clockwise rotation: \((x, y)\) becomes \((y, -x)\)? Wait, no, let's take point \((2, -6)\). So \(x = 2\), \(y = -6\). Then \(y = -6\), \(-x = -2\), so \((-6, -2)\), which is the second option? Wait, the second option is \(H'(-6, -2)\). Wait, but let's verify with the rotation matrix. The rotation matrix for 90° clockwise is \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\). So multiplying \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\) with \(\begin{pmatrix}2\\-6\end{pmatrix}\) gives \(\begin{pmatrix}0*2 + 1*(-6)\\-1*2 + 0*(-6)\end{pmatrix}=\begin{pmatrix}-6\\-2\end{pmatrix}\), which is \((-6, -2)\), which is the second option. Wait, but let's check the options again. The options are:

1. \(H'(6, 2)\)

2. \(H'(-6, -2)\)

3. \(H'(6, -2)\)

4. \(H'(-6, 2)\)

So according to the calculation, the image is \((-6, -2)\), which is the second option.

Answer:

B. \(H'(-6, -2)\) (assuming the options are labeled as A, B, C, D with A: \(H'(6, 2)\), B: \(H'(-6, -2)\), C: \(H'(6, -2)\), D: \(H'(-6, 2)\))