Question

a square has a vertex at (2, -3). after a 270-degree counterclockwise rotation around the origin, what is the image of this vertex?

a. (3,2)

b. (-3,-2)

c. (-2,3)

d. (3,-2)

Explanation:

Step1: Recall rotation rule

For a 270° counterclockwise rotation about the origin, the rule is \((x,y)\to(y,-x)\).

Step2: Apply the rule to \((2,-3)\)

Here, \(x = 2\) and \(y=-3\). Substitute into the rule: \(y=-3\), \(-x=-2\)? Wait, no, wait: Wait, the correct rule for 270° counterclockwise (or 90° clockwise) is \((x,y)\to(y, -x)\). Wait, let's recheck:
Wait, standard rotation rules:

• 90° counterclockwise: \((x,y)\to(-y,x)\)
• 180° counterclockwise: \((x,y)\to(-x,-y)\)
• 270° counterclockwise: \((x,y)\to(y, -x)\) (because 270° counterclockwise is equivalent to 90° clockwise, and the rule for 90° clockwise is \((x,y)\to(y, -x)\))

So for point \((2, -3)\), \(x = 2\), \(y=-3\). Applying the 270° counterclockwise rule: \((x,y)\to(y, -x)\) becomes \((-3, -2)\)? Wait, no, wait: Wait, \(y\) is \(-3\), \(-x\) is \(-2\)? Wait, no, \(x = 2\), so \(-x=-2\)? Wait, no, wait, let's do it again. Wait, maybe I mixed up the rule. Let's use the rotation matrix. The rotation matrix for \(\theta\) degrees counterclockwise is \(

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

\). For \(\theta = 270^\circ\), \(\cos270^\circ = 0\), \(\sin270^\circ=-1\). So the matrix is \(

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

\) (because \(-\sin270^\circ = 1\), \(\sin270^\circ=-1\), \(\cos270^\circ = 0\)). So multiplying by vector \(

$$\begin{pmatrix}x\\y\end{pmatrix}$$

=

$$\begin{pmatrix}2\\-3\end{pmatrix}$$

\):

\(

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

$$\begin{pmatrix}2\\-3\end{pmatrix}$$

=

$$\begin{pmatrix}0\times2 + 1\times(-3)\\-1\times2+0\times(-3)\end{pmatrix}$$

=

$$\begin{pmatrix}-3\\-2\end{pmatrix}$$

\). Wait, but that's option b? Wait, but let's check again. Wait, maybe I made a mistake in the rotation matrix. Wait, 270° counterclockwise: the rotation matrix is \(

$$\begin{pmatrix}\cos270&-\sin270\\\sin270&\cos270\end{pmatrix}$$

=

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

\) (since \(\cos270 = 0\), \(\sin270=-1\), so \(-\sin270 = 1\), \(\sin270=-1\)). So applying to (2, -3):

First component: \(0\times2 + 1\times(-3)= -3\)

Second component: \(-1\times2 + 0\times(-3)= -2\)

So the image is (-3, -2), which is option b. Wait, but let's check the rule again. Alternatively, 270° counterclockwise rotation is the same as 90° clockwise rotation. The rule for 90° clockwise rotation is (x, y) → (y, -x). So for (2, -3), x=2, y=-3. So (y, -x) = (-3, -2). Yes, that matches. So the image is (-3, -2), which is option b.

Answer:

b. \((-3, -2)\)