Question

triangle xyz is rotated to create the image triangle xyz. which rules could describe the rotation? select two options. \\(\square r_{0, 90^{\circ}}\\) \\(\square r_{0, 180^{\circ}}\\) \\(\square r_{0, 270^{\circ}}\\) \\(\square (x, y) \to (-y, x)\\) \\(\square (x, y) \to (-x, -y)\\)

Explanation:

Step 1: Identify Coordinates

First, find coordinates of original and image points. Let's take point \( Z(0, 4) \), \( X(-2, 2) \), \( Y(2, 2) \) and their images \( Z'(0, -4) \), \( X'(2, -2) \), \( Y'(-2, -2) \).

Step 2: Analyze Rotation Rules

• \( R_{0, 180^\circ} \): Rule is \( (x, y) \to (-x, -y) \). Check \( Z(0,4) \to (0, -4) \) (matches \( Z' \)), \( X(-2,2) \to (2, -2) \) (matches \( X' \)), \( Y(2,2) \to (-2, -2) \) (matches \( Y' \)). So \( R_{0, 180^\circ} \) and \( (x, y) \to (-x, -y) \) work.
• \( R_{0, 90^\circ} \): Rule \( (x, y) \to (-y, x) \). For \( Z(0,4) \to (-4, 0) \) (not \( Z' \)), so no.
• \( R_{0, 270^\circ} \): Rule \( (x, y) \to (y, -x) \). For \( Z(0,4) \to (4, 0) \) (not \( Z' \)), so no.
• \( (x, y) \to (-y, x) \): As above, doesn't match \( Z \)'s image, so no.
• \( (x, y) \to (-x, -y) \): Matches all points as shown.

Answer:

\( R_{0, 180^\circ} \), \( (x, y) \to (-x, -y) \)