Question

one vertex of a polygon is located at (3, - 2). after a rotation, the vertex is located at (2, 8). which transformations could have taken place? choose two correct answers

Explanation:

Step1: Recall rotation rules about the origin

For a point $(x,y)$ rotated $90^{\circ}$ counter - clockwise about the origin $(0,0)$, the new point is $(-y,x)$. For a $180^{\circ}$ rotation about the origin, the new point is $(-x,-y)$. For a $270^{\circ}$ counter - clockwise (or $- 90^{\circ}$) rotation about the origin, the new point is $(y,-x)$.

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

If we rotate the point $(3,-2)$ $90^{\circ}$ counter - clockwise about the origin ($R_{0,90^{\circ}}$), the new point is $-(-2),3=(2,3)$.
If we rotate the point $(3,-2)$ $180^{\circ}$ about the origin ($R_{0,180^{\circ}}$), the new point is $(-3,2)$.
If we rotate the point $(3,-2)$ $270^{\circ}$ counter - clockwise about the origin ($R_{0,270^{\circ}}$), the new point is $(-2,-3)$.
If we rotate the point $(3,-2)$ $-90^{\circ}$ (or $270^{\circ}$ clockwise) about the origin ($R_{0,-90^{\circ}}$), the new point is $(-2, - 3)$.
If we rotate the point $(3,-2)$ $-170^{\circ}$ is not a standard rotation we usually consider in basic geometry.
Let's use the rotation matrix for a rotation of $\theta$ about the origin. The rotation matrix for a counter - clockwise rotation of angle $\theta$ is

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

.
For a point $(x,y)$ written as a column vector

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

, the new point

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

.
For $\theta = 90^{\circ}$, $\cos90^{\circ}=0,\sin90^{\circ}=1$, and for the point $(3,-2)$:

$$\begin{bmatrix}0&- 1\\1&0\end{bmatrix}\begin{bmatrix}3\\-2\end{bmatrix}=\begin{bmatrix}2\\3\end{bmatrix}$$

.
For $\theta=-90^{\circ}$, $\cos(-90^{\circ}) = 0,\sin(-90^{\circ})=-1$, and

$$\begin{bmatrix}0&1\\-1&0\end{bmatrix}\begin{bmatrix}3\\-2\end{bmatrix}=\begin{bmatrix}-2\\-3\end{bmatrix}$$

.

Answer:

There are no correct answers among the given options.