Question

what are the vertices for the final image after applying the composition $t_{-2,4} \circ r_{o,180\degree}$ to $\triangle xyz$?
$x$ is
$y$ is
$z$ is

Explanation:

First, we need to identify the original coordinates of \( \triangle XYZ \). From the graph, we can see:

• \( X(2, 5) \) (assuming the grid lines, let's confirm: looking at the y-axis, X is at y=5? Wait, maybe I misread. Wait, the original triangle: Y is at (2,3)? Wait, no, let's check the grid. Let's re-express:

Wait, the original \( \triangle XYZ \): Let's find coordinates. Let's assume each grid square is 1 unit. So:

• \( X \): x=2, y=5? Wait, no, the y-axis has 4 at the top, so maybe X is (2,5)? Wait, no, the green triangle: Y is at (2,3), Z at (4,3), X at (2,5). So original coordinates:
• \( X(2, 5) \)
• \( Y(2, 3) \)
• \( Z(4, 3) \)

Now, the composition is \( T_{-2, 4} \circ R_{O, 180^\circ} \), which means we first apply the rotation \( R_{O, 180^\circ} \) (rotation about the origin by 180 degrees) and then the translation \( T_{-2, 4} \) (translate 2 units left and 4 units up).

Step 1: Apply \( R_{O, 180^\circ} \) to each vertex.

The rule for a 180-degree rotation about the origin is \( (x, y) \to (-x, -y) \).

• For \( X(2, 5) \): \( X' = (-2, -5) \)
• For \( Y(2, 3) \): \( Y' = (-2, -3) \)
• For \( Z(4, 3) \): \( Z' = (-4, -3) \)

Step 2: Apply translation \( T_{-2, 4} \) (which is \( (x, y) \to (x - 2, y + 4) \)) to each rotated vertex.

• For \( X'(-2, -5) \): \( X'' = (-2 - 2, -5 + 4) = (-4, -1) \)
• For \( Y'(-2, -3) \): \( Y'' = (-2 - 2, -3 + 4) = (-4, 1) \)
• For \( Z'(-4, -3) \): \( Z'' = (-4 - 2, -3 + 4) = (-6, 1) \)

Wait, but let's check the original coordinates again. Maybe I made a mistake in original coordinates. Let's re-examine the graph:

Looking at the grid, the green triangle: Y is at (2, 3), Z at (4, 3), X at (2, 5). So original coordinates:

• \( X(2, 5) \)
• \( Y(2, 3) \)
• \( Z(4, 3) \)

Rotation 180 degrees: \( (x,y) \to (-x, -y) \)

So \( X'(-2, -5) \), \( Y'(-2, -3) \), \( Z'(-4, -3) \)

Then translation \( T_{-2, 4} \): \( (x - 2, y + 4) \)

So:

• \( X'' \): \( -2 - 2 = -4 \), \( -5 + 4 = -1 \) → \( (-4, -1) \)
• \( Y'' \): \( -2 - 2 = -4 \), \( -3 + 4 = 1 \) → \( (-4, 1) \)
• \( Z'' \): \( -4 - 2 = -6 \), \( -3 + 4 = 1 \) → \( (-6, 1) \)

Wait, but let's confirm with the graph's lower triangle (Z', Y', X'). Let's see Z' is at (-4, -3)? Wait, no, the lower triangle: Z' is at (-4, -3)? Wait, maybe my original coordinates are wrong. Let's re-express the original triangle:

Looking at the x-axis: X is at x=2, Y at x=2, Z at x=4. Y and Z are at y=3, X at y=5. So original:

\( X(2, 5) \), \( Y(2, 3) \), \( Z(4, 3) \)

Rotation 180: \( X'(-2, -5) \), \( Y'(-2, -3) \), \( Z'(-4, -3) \)

Translation \( T_{-2, 4} \): (x - 2, y + 4)

So:

\( X'' = (-2 - 2, -5 + 4) = (-4, -1) \)

\( Y'' = (-2 - 2, -3 + 4) = (-4, 1) \)

\( Z'' = (-4 - 2, -3 + 4) = (-6, 1) \)

So the vertices after composition are:

\( X''(-4, -1) \), \( Y''(-4, 1) \), \( Z''(-6, 1) \)

Answer:

\( X'' \) is \((-4, -1)\)
\( Y'' \) is \((-4, 1)\)
\( Z'' \) is \((-6, 1)\)