Question

determine the angles of rotation. choose all answers that apply: a $90^{circ}$ clockwise b $90^{circ}$ counterclockwise c $180^{circ}$ d $270^{circ}$ clockwise

Explanation:

Step1: Recall rotation rules

For a point \((x,y)\):
- \(90^\circ\) clockwise: \((y, -x)\)
- \(90^\circ\) counterclockwise: \((-y, x)\)
- \(180^\circ\): \((-x, -y)\)
- \(270^\circ\) clockwise (or \(90^\circ\) counterclockwise): \((-y, x)\) Wait, no: \(270^\circ\) clockwise is equivalent to \(90^\circ\) counterclockwise? Wait, correction: \(270^\circ\) clockwise rotation of \((x,y)\) is \((-y, x)\)? No, let's re - establish:
The correct rotation rules:
- Rotation of \(90^\circ\) clockwise about the origin: \((x,y)\to(y, -x)\)
- Rotation of \(90^\circ\) counter - clockwise about the origin: \((x,y)\to(-y, x)\)
- Rotation of \(180^\circ\) about the origin: \((x,y)\to(-x, -y)\)
- Rotation of \(270^\circ\) clockwise about the origin: \((x,y)\to(-y, x)\) (which is the same as \(90^\circ\) counter - clockwise)
- Rotation of \(270^\circ\) counter - clockwise about the origin: \((x,y)\to(y, -x)\) (which is the same as \(90^\circ\) clockwise)

Let's take a point from the blue figure (pre - image) and its corresponding point in the red figure (image). Let's assume the center of rotation is the origin (we can check the coordinates). Let's find the coordinates of a point, say point \(D\) (blue) and \(D\) (red)? Wait, looking at the graph, let's take point \(C\) (blue) and \(C'\) (red). Wait, maybe better to take a point with clear coordinates. Let's assume the blue figure has a point \(A\) at \((-1,-6)\) (approximate from the grid) and the red figure has \(A'\) at \((6, - 2)\)? Wait, maybe I made a mistake. Alternatively, let's think about the orientation.

Wait, another approach: The rotation from the blue to the red figure. Let's consider the direction. If we rotate \(90^\circ\) counter - clockwise or \(270^\circ\) clockwise, the orientation changes in a way that matches. Wait, let's check the angle of rotation.

Wait, the key is to see the relationship between the pre - image and the image. Let's consider the rotation of \(90^\circ\) counter - clockwise or \(270^\circ\) clockwise.

Wait, let's take a point. Let's suppose the center of rotation is the origin. Let's take a point from the blue shape, say point \(B\) (blue) with coordinates (let's estimate from the grid) \((-3,-6)\) and the corresponding point \(B'\) (red) with coordinates \((6,3)\)? Wait, no, maybe the center of rotation is not the origin. Wait, maybe the figure is rotated about a point, but the general idea is to check the angle.

Wait, the correct answers are \(90^\circ\) counterclockwise (option B) and \(270^\circ\) clockwise (option D) because \(90^\circ\) counterclockwise rotation and \(270^\circ\) clockwise rotation are equivalent (since \(360 - 90=270\)). Let's verify with a point. Suppose we have a point \((x,y)\). After \(90^\circ\) counterclockwise rotation, it becomes \((-y,x)\). After \(270^\circ\) clockwise rotation, the formula is \((x,y)\to(-y,x)\) (since \(270^\circ\) clockwise rotation matrix is \(\begin{pmatrix}0&1\\ - 1&0\end{pmatrix}\) which gives \((y,-x)\)? Wait, I think I messed up the rotation matrices.

The rotation matrix for \(90^\circ\) clockwise is \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\), so \((x,y)\begin{pmatrix}0&1\\-1&0\end{pmatrix}=(y, - x)\)

The rotation matrix for \(90^\circ\) counter - clockwise is \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\), so \((x,y)\begin{pmatrix}0&-1\\1&0\end{pmatrix}=(-y, x)\)

The rotation matrix for \(180^\circ\) is \(\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\), so \((x,y)\begin{pmatrix}-1&0\\0&-1\end{pmatrix}=(-x, - y)\)

The rotation matrix for \(270^\circ\) clockwise is the same as \(90^\circ\) counter - clockwise, so the matrix is \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\), so \((x,y)\to(-y, x)\)

The rotation matrix for \(270^\circ\) counter - clockwise is the same as \(90^\circ\) clockwise, so the matrix is \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\), so \((x,y)\to(y, - x)\)

Now, let's take a point from the blue figure (pre - image) and the red figure (image). Let's assume the center of rotation is the origin. Let's pick a point, say, if we consider the blue figure's point \(C\) (let's say coordinates \((-3,-3)\)) and the red figure's point \(C'\) (let's say coordinates \((3,3)\))? No, that would be \(180^\circ\), but that's not the case. Wait, maybe my coordinate estimation is wrong.

Wait, looking at the graph, the blue figure is on the left - hand side (negative \(x\) - axis side) and the red figure is on the right - hand side (positive \(x\) - axis side). Let's take a point from the blue figure, say a vertex, and its image. Let's suppose we rotate \(90^\circ\) counterclockwise or \(270^\circ\) clockwise.

If we rotate a figure \(90^\circ\) counterclockwise or \(270^\circ\) clockwise, the orientation of the figure changes in a way that matches the transformation from blue to red.

Let's consider the direction of rotation. A \(90^\circ\) counterclockwise rotation and a \(270^\circ\) clockwise rotation are equivalent in terms of the final position (since \(360 - 90=270\), and rotating \(270^\circ\) clockwise is the same as rotating \(90^\circ\) counterclockwise).

So the applicable rotations are \(90^\circ\) counterclockwise (option B) and \(270^\circ\) clockwise (option D).

Answer:

B. \(90^\circ\) counterclockwise, D. \(270^\circ\) clockwise