Triangle XYZ has vertices at X(−5, −5), Y(2, 2), Z(2, −4).
Part A: If triangle XYZ is rotated 90° counterclockwise about the origin, determine the coordinates of the vertices for the image, triangle X′Y′Z′. Show each step of your work. (4 points)
Part B: If triangle XYZ is rotated 270° clockwise about the origin, determine the coordinates of the vertices for the image, triangle X′Y′Z′. Show each step of your work. (4 points)
Part C: Describe the similarities and differences between the two rotations in parts A and B. (4 points)
Part A:
To rotate triangle XYZ 90° counterclockwise about the origin, we can use the following rotation matrix:
\[
\begin{bmatrix}
0 & -1 \\
1 & 0 \\
\end{bmatrix}
\]
Applying this matrix to each of the vertices of triangle XYZ:
For vertex X(-5, -5):
\[
\begin{bmatrix}
0 & -1 \\
1 & 0 \\
\end{bmatrix}
\begin{bmatrix}
-5 \\
-5 \\
\end{bmatrix}
=
\begin{bmatrix}
5 \\
-5 \\
\end{bmatrix}
\]
So the new coordinates for X' are (5, -5).
For vertex Y(2, 2):
\[
\begin{bmatrix}
0 & -1 \\
1 & 0 \\
\end{bmatrix}
\begin{bmatrix}
2 \\
2 \\
\end{bmatrix}
=
\begin{bmatrix}
-2 \\
2 \\
\end{bmatrix}
\]
So the new coordinates for Y' are (-2, 2).
For vertex Z(2, -4):
\[
\begin{bmatrix}
0 & -1 \\
1 & 0 \\
\end{bmatrix}
\begin{bmatrix}
2 \\
-4 \\
\end{bmatrix}
=
\begin{bmatrix}
4 \\
2 \\
\end{bmatrix}
\]
So the new coordinates for Z' are (4, 2).
Therefore, the image of triangle XYZ after a 90° counterclockwise rotation about the origin is X'(5, -5), Y'(-2, 2), Z'(4, 2).
Part B:
To rotate triangle XYZ 270° clockwise about the origin, we can use the following rotation matrix:
\[
\begin{bmatrix}
0 & 1 \\
-1 & 0 \\
\end{bmatrix}
\]
Applying this matrix to each of the vertices of triangle XYZ:
For vertex X(-5, -5):
\[
\begin{bmatrix}
0 & 1 \\
-1 & 0 \\
\end{bmatrix}
\begin{bmatrix}
-5 \\
-5 \\
\end{bmatrix}
=
\begin{bmatrix}
5 \\
-5 \\
\end{bmatrix}
\]
So the new coordinates for X' are (5, -5).
For vertex Y(2, 2):
\[
\begin{bmatrix}
0 & 1 \\
-1 & 0 \\
\end{bmatrix}
\begin{bmatrix}
2 \\
2 \\
\end{bmatrix}
=
\begin{bmatrix}
-2 \\
-2 \\
\end{bmatrix}
\]
So the new coordinates for Y' are (-2, -2).
For vertex Z(2, -4):
\[
\begin{bmatrix}
0 & 1 \\
-1 & 0 \\
\end{bmatrix}
\begin{bmatrix}
2 \\
-4 \\
\end{bmatrix}
=
\begin{bmatrix}
4 \\
-2 \\
\end{bmatrix}
\]
So the new coordinates for Z' are (4, -2).
Therefore, the image of triangle XYZ after a 270° clockwise rotation about the origin is X'(5, -5), Y'(-2, -2), Z'(4, -2).
Part C:
The similarities between the two rotations in parts A and B are that both operations result in a triangle with the same orientation, as they both involve rotating the original triangle about the origin. Additionally, both rotations result in the same overall shape of the triangle, just in different positions.
The main difference between the two rotations is the direction in which the rotation is performed and the final positions of the vertices. In part A, the triangle is rotated 90° counterclockwise, while in part B, the triangle is rotated 270° clockwise. This results in the vertices of the image triangle being different in both cases.