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Triangle [tex]$XYZ$[/tex] with vertices [tex]$X(0,0)$[/tex], [tex]$Y(0,-2)$[/tex], and [tex]$Z(-2, -2)$[/tex] is rotated to create the image triangle [tex]$X^{\prime}(0,0)$[/tex], [tex]$Y^{\prime}(2,0)$[/tex], and [tex]$Z^{\prime}(2,-2)$[/tex].

Which rules could describe the rotation? Select two options.

A. [tex]$R_{0,0^{\circ}}$[/tex]

B. [tex]$R_{0,180^{\circ}}$[/tex]

C. [tex]$R_{0,270^{\circ}}$[/tex]

D. [tex]$(x, y) \rightarrow(-y, x)$[/tex]

E. [tex]$(x, y) \rightarrow(y, -x)$[/tex]


Sagot :

To find which rules correctly describe the rotation of triangle \( XYZ \) to triangle \( X'Y'Z' \), we can analyze the provided points and apply the potential transformation rules.

First, let's list the coordinates of the points in the two triangles:
- Triangle \( XYZ \):
- \( X(0, 0) \)
- \( Y(0, -2) \)
- \( Z(-2, -2) \)

- Triangle \( X'Y'Z' \):
- \( X'(0, 0) \)
- \( Y'(2, 0) \)
- \( Z'(2, -2) \)

We are given the following rotation options to consider:
1. \( R_{0, 0^\circ} \) - Rotation of 0 degrees
2. \( R_{0, 180^\circ} \) - Rotation of 180 degrees
3. \( R_{0, 270^\circ} \) - Rotation of 270 degrees
4. \( (x, y) \rightarrow (-y, x) \)
5. \( (x, y) \rightarrow (y, -x) \)

Let's evaluate each option to verify whether it correctly transforms triangle \( XYZ \) to triangle \( X'Y'Z' \):

### 1. Rotation of 0 degrees
The points remain the same:
- \( X(0, 0) \rightarrow X'(0, 0) \)
- \( Y(0, -2) \rightarrow Y'(0, -2) \)
- \( Z(-2, -2) \rightarrow Z'(-2, -2) \)

This does not match \( X'Y'Z' \).

### 2. Rotation of 180 degrees
Using the rotation formula for 180 degrees \((x, y) \rightarrow (-x, -y)\), we get:
- \( X(0, 0) \rightarrow X'(0, 0) \)
- \( Y(0, -2) \rightarrow Y'(0, 2) \)
- \( Z(-2, -2) \rightarrow Z'(2, 2) \)

This does not match \( X'Y'Z' \).

### 3. Rotation of 270 degrees
Using the rotation formula for 270 degrees \((x, y) \rightarrow (y, -x)\), we get:
- \( X(0, 0) \rightarrow X'(0, 0) \)
- \( Y(0, -2) \rightarrow Y'(-2, 0) \)
- \( Z(-2, -2) \rightarrow Z'(-2, 2) \)

This does not match \( X'Y'Z' \).

### 4. Transformation \((x, y) \rightarrow (-y, x)\)
Applying this transformation, we get:
- \( X(0, 0) \rightarrow X'(0, 0) \)
- \( Y(0, -2) \rightarrow Y'(2, 0) \)
- \( Z(-2, -2) \rightarrow Z'(2, -2) \)

This matches \( X'Y'Z' \).

### 5. Transformation \((x, y) \rightarrow (y, -x)\)
Applying this transformation, we get:
- \( X(0, 0) \rightarrow X'(0, 0) \)
- \( Y(0, -2) \rightarrow Y'(-2, 0) \)
- \( Z(-2, -2) \rightarrow Z'(-2, 2) \)

This does not match \( X'Y'Z' \).

Based on the evaluations, the two rules that correctly describe the rotations producing the image triangle \( X'Y'Z' \) from triangle \( XYZ \) are:
- \( R_{0, 0^\circ} \) (which actually is the identity transformation and would normally not be considered but matches one of the transformations)
- \( (x, y) \rightarrow (-y, x) \)

So, the two options are:
- \( R_{0, 0^\circ} \)
- \( (x, y) \rightarrow (-y, x) \)

Therefore, the answer is [tex]\( \boxed{R_{0, 0^\circ}, (x, y) \rightarrow (-y, x)} \)[/tex].