Question

this graph shows a rhombus with its center at the origin.
select all of the transformations that map the rhombus onto itself.

• reflection across the line $y = x$
• $180^circ$ rotation about the origin followed by reflection across the $x$-axis
• $90^circ$ counterclockwise rotation about the origin
• reflection across the line $y = 2$ followed by translation 4 units down

Answer:

To solve this, we analyze each transformation:

1. Reflection across \( y = x \)

• The rhombus has vertices (let's identify them: from the graph, vertices are \((-4,0)\), \((0,2)\), \((4,0)\), \((0,-2)\)). Reflecting \((-4,0)\) over \( y = x \) gives \((0,-4)\), which is not a vertex of the rhombus. So this transformation does not map the rhombus onto itself.

2. \( 180^\circ \) rotation about the origin followed by reflection across the \( x \)-axis

• First, \( 180^\circ \) rotation: A point \((x,y)\) becomes \((-x,-y)\). So \((-4,0) \to (4,0)\), \((0,2) \to (0,-2)\), \((4,0) \to (-4,0)\), \((0,-2) \to (0,2)\).
• Then reflect across \( x \)-axis: \((x,y) \to (x,-y)\). So \((4,0) \to (4,0)\), \((0,-2) \to (0,2)\), \((-4,0) \to (-4,0)\), \((0,2) \to (0,-2)\). This brings us back to the original vertices. So this transformation does map the rhombus onto itself.

3. \( 90^\circ \) counterclockwise rotation about the origin

• A \( 90^\circ \) counterclockwise rotation transforms \((x,y) \to (-y,x)\). For \((-4,0)\): \((-0,-4) = (0,-4)\) (not a vertex). So this transformation does not map the rhombus onto itself.

4. Reflection across \( y = 2 \) followed by translation 4 units down

• Reflect across \( y = 2 \): The distance from a point \((x,y)\) to \( y = 2 \) is \( |y - 2| \), so the reflected point is \((x, 2 + (2 - y)) = (x, 4 - y)\).
• For \((-4,0)\): \( ( -4, 4 - 0 ) = (-4,4) \).
• Then translate 4 units down: \( (-4,4 - 4) = (-4,0) \) (original vertex).
• For \((0,2)\): \( (0, 4 - 2 ) = (0,2) \), then translate 4 down: \( (0, -2) \) (original vertex).
• For \((4,0)\): \( (4, 4 - 0 ) = (4,4) \), translate 4 down: \( (4,0) \) (original vertex).
• For \((0,-2)\): \( (0, 4 - (-2)) = (0,6) \), translate 4 down: \( (0,2) \) (original vertex). So this transformation does map the rhombus onto itself. Wait, no—wait, let's recheck. Wait, the original vertices are \((-4,0)\), \((0,2)\), \((4,0)\), \((0,-2)\). Let's do each step:
• Reflect \((-4,0)\) over \( y=2 \): distance to \( y=2 \) is \( 2 - 0 = 2 \), so reflected \( y \)-coordinate is \( 2 + 2 = 4 \), so \((-4,4)\). Then translate 4 down: \((-4, 4 - 4) = (-4,0)\) (good).
• Reflect \((0,2)\) over \( y=2 \): \( (0,2) \) (since it's on the line), then translate 4 down: \( (0, -2) \) (good, original vertex).
• Reflect \((4,0)\) over \( y=2 \): \( (4,4) \), translate 4 down: \( (4,0) \) (good).
• Reflect \((0,-2)\) over \( y=2 \): distance is \( 2 - (-2) = 4 \), so reflected \( y \)-coordinate is \( 2 + 4 = 6 \), translate 4 down: \( (0, 6 - 4) = (0,2) \) (good, original vertex). Wait, but the problem says "map the rhombus onto itself"—does this transformation result in the same rhombus? Let's see the order: reflect then translate. After reflection, the points are \((-4,4)\), \((0,2)\), \((4,4)\), \((0,6)\). Then translate 4 down: \((-4,0)\), \((0,-2)\), \((4,0)\), \((0,2)\)—which is the original rhombus. So this works? Wait, but maybe I made a mistake earlier. Wait, no—wait, the original rhombus has vertices at \((-4,0)\), \((0,2)\), \((4,0)\), \((0,-2)\). After reflection over \( y=2 \) and translation 4 down, we get back to these vertices. So this transformation works? Wait, but let's check the other options again.

Wait, maybe my initial analysis of the first option was wrong. Let's re-express the vertices clearly:

From the graph, the rhombus has vertices:

• Left: \((-4, 0)\)
• Top: \((0, 2)\)
• Right: \((4, 0)\)
• Bottom: \((0, -2)\)

Re-analyzing each option:

1. Reflection across \( y = x \):

The reflection of \((-4, 0)\) over \( y = x \) is \((0, -4)\) (not a vertex).
The reflection of \((0, 2)\) over \( y = x \) is \((2, 0)\) (not a vertex).
So this transformation fails.

1. \( 180^\circ \) rotation about the origin followed by reflection across the \( x \)-axis:

• \( 180^\circ \) rotation: \((x, y) \to (-x, -y)\).
• \((-4, 0) \to (4, 0)\)
• \((0, 2) \to (0, -2)\)
• \((4, 0) \to (-4, 0)\)
• \((0, -2) \to (0, 2)\)
• Then reflect across \( x \)-axis: \((x, y) \to (x, -y)\).
• \((4, 0) \to (4, 0)\)
• \((0, -2) \to (0, 2)\)
• \((-4, 0) \to (-4, 0)\)
• \((0, 2) \to (0, -2)\)

This returns to the original vertices. So this transformation works.

1. \( 90^\circ \) counterclockwise rotation about the origin:

\( 90^\circ \) counterclockwise rotation: \((x, y) \to (-y, x)\).

• \((-4, 0) \to (0, -4)\) (not a vertex).
• \((0, 2) \to (-2, 0)\) (not a vertex).

So this transformation fails.

1. Reflection across \( y = 2 \) followed by translation 4 units down:

• Reflect across \( y = 2 \): For a point \((x, y)\), the reflected \( y \)-coordinate is \( 2 + (2 - y) = 4 - y \).
• \((-4, 0)\): \( y = 0 \to 4 - 0 = 4 \), so \((-4, 4)\).
• \((0, 2)\): \( y = 2 \to 4 - 2 = 2 \), so \((0, 2)\).
• \((4, 0)\): \( y = 0 \to 4 - 0 = 4 \), so \((4, 4)\).
• \((0, -2)\): \( y = -2 \to 4 - (-2) = 6 \), so \((0, 6)\).
• Translate 4 units down: Subtract 4 from \( y \)-coordinate.
• \((-4, 4) \to (-4, 0)\) (original vertex).
• \((0, 2) \to (0, -2)\) (original vertex).
• \((4, 4) \to (4, 0)\) (original vertex).
• \((0, 6) \to (0, 2)\) (original vertex).

This returns to the original vertices. So this transformation works.

Wait, but the initial check for the first option was wrong, and the fourth option works. But let's confirm the problem: "Select all of the transformations that map the rhombus onto itself."

Wait, maybe I made a mistake with the \( 180^\circ \) rotation + reflection. Let's re-express:

After \( 180^\circ \) rotation, the vertices are \((4, 0)\), \((0, -2)\), \((-4, 0)\), \((0, 2)\). Then reflecting across the \( x \)-axis: \((4, 0) \to (4, 0)\), \((0, -2) \to (0, 2)\), \((-4, 0) \to (-4, 0)\), \((0, 2) \to (0, -2)\). Which is the original set of vertices (just reordered). So that works.

For the fourth option: reflection over \( y=2 \) and translation 4 down. After reflection, the top and bottom vertices swap in a way that translation brings them back. So that works.

Wait, but the original check for the first option (reflection over \( y=x \)): the reflected points are not vertices, so it fails. The \( 90^\circ \) rotation fails. So the correct transformations are:

• \( 180^\circ \) rotation about the origin followed by reflection across the \( x \)-axis
• Reflection across the line \( y = 2 \) followed by translation 4 units down

But wait, let's check the fourth option again. The original rhombus has a vertical axis of symmetry (through \( x=0 \)) and horizontal axis (through \( y=0 \))? No, the diagonals are along the \( x \)-axis (length 8) and \( y \)-axis (length 4). So reflecting over \( y=2 \) (which is 2 units above the top vertex \((0,2)\)): the distance from \((0,2)\) to \( y=2 \) is 0, so it stays. The distance from \((0,-2)\) to \( y=2 \) is 4, so reflected to \( y=2 + 4 = 6 \), then translating 4 down gives \( y=2 \), which is the top vertex. Wait, no—wait, \((0,-2)\) reflected over \( y=2 \) is \( (0, 2 + (2 - (-2))) = (0, 6) \), then translating 4 down: \( (0, 6 - 4) = (0, 2) \) (top vertex). The bottom vertex \((0,-2)\) becomes the top vertex, and the top vertex \((0,2)\) becomes the bottom vertex \((0,-2)\) after translation? Wait, no:

Wait, original vertices:
\( A(-4, 0) \), \( B(0, 2) \), \( C(4, 0) \), \( D(0, -2) \)

After reflection over \( y=2 \):

• \( A(-4, 0) \): distance to \( y=2 \) is \( 2 - 0 = 2 \), so reflected \( y = 2 + 2 = 4 \) → \( A'(-4, 4) \)
• \( B(0, 2) \): distance to \( y=2 \) is 0 → \( B'(0, 2) \)
• \( C(4, 0) \): distance to \( y=2 \) is \( 2 - 0 = 2 \) → reflected \( y = 2 + 2 = 4 \) → \( C'(4, 4) \)
• \( D(0, -2) \): distance to \( y=2 \) is \( 2 - (-2) = 4 \) → reflected \( y = 2 + 4 = 6 \) → \( D'(0, 6) \)

Then translate 4 units down:

• \( A'(-4, 4) \to (-4, 0) = A \)
• \( B'(0, 2) \to (0, -2) = D \)
• \( C'(4, 4) \to (4, 0) = C \)
• \( D'(0, 6) \to (0, 2) = B \)

So the transformed vertices are \( A, D, C, B \), which is the original rhombus (just the order of vertices changed, but the set is the same). So this transformation works.

Final Answer:

The correct transformations are:

• \( 180^\circ \) rotation about the origin followed by reflection across the \( x \)-axis
• Reflection across the line \( y = 2 \) followed by translation 4 units down

(Note: The initial check for the first option was incorrect; it fails. The \( 90^\circ \) rotation also fails.)