Question

one vertex of a polygon is located at (3, -2). after a rotation, the vertex is located at (2, 3). which transformations could have taken place? choose two correct answers.

Explanation:

Step1: Recall rotation rules about the origin

The rotation matrix for a counter - clockwise rotation of a point $(x,y)$ about the origin by an angle $\theta$ is given by

$$\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$$

, where $(x',y')$ is the new point after rotation. For a $90^{\circ}$ counter - clockwise rotation ($\theta = 90^{\circ}$, $\cos90^{\circ}=0$, $\sin90^{\circ}=1$), the transformation is $(x,y)\to(-y,x)$. For a $270^{\circ}$ counter - clockwise rotation ($\theta = 270^{\circ}$, $\cos270^{\circ}=0$, $\sin270^{\circ}=- 1$), the transformation is $(x,y)\to(y, - x)$. For a $-90^{\circ}$ rotation (which is a $270^{\circ}$ counter - clockwise rotation) and a $-270^{\circ}$ rotation (which is a $90^{\circ}$ counter - clockwise rotation).

Step2: Apply rotation rules to the point $(3,-2)$

For a $90^{\circ}$ counter - clockwise rotation ($R_{0,90^{\circ}}$) of the point $(3,-2)$: Using the rule $(x,y)\to(-y,x)$, we substitute $x = 3$ and $y=-2$. Then $(-y,x)=-(-2),3=(2,3)$.
For a $- 270^{\circ}$ rotation ($R_{0,-270^{\circ}}$) which is equivalent to a $90^{\circ}$ counter - clockwise rotation, applying the rule $(x,y)\to(-y,x)$ to the point $(3,-2)$ also gives $(2,3)$.

Answer:

B. $R_{0,90^{\circ}}$
E. $R_{0,-270^{\circ}}$