Question

which transformation does not produce a square with vertices at the same set of coordinate pairs? a square is shown on the coordinate grid. the square has vertices at (-2, 2), (2, 2), (2, -2), (-2, -2). the transformation options are: (x, y) → (-x, y); (x, y) → (x - 2, y - 2); (x, y) → (-x, -y); (x, y) → (-y, x)

Explanation:

Step1: Analyze each transformation

First, let's recall the original square's vertices: \((-2, 2)\), \((2, 2)\), \((2, -2)\), \((-2, -2)\).

Transformation 1: \((x, y) \to (-x, y)\)

Apply to each vertex:
- \((-2, 2) \to (2, 2)\)
- \((2, 2) \to (-2, 2)\)
- \((2, -2) \to (-2, -2)\)
- \((-2, -2) \to (2, -2)\)
These are the same vertices (just reordered), so it produces the same square.

Transformation 2: \((x, y) \to (x - 2, y - 2)\)

Apply to each vertex:
- \((-2, 2) \to (-4, 0)\)
- \((2, 2) \to (0, 0)\)
- \((2, -2) \to (0, -4)\)
- \((-2, -2) \to (-4, -4)\)
This is a square, but with different coordinates (translated), not the same set of coordinate pairs as original. Wait, no—wait, the question is which does NOT produce a square with the same set of coordinate pairs. Wait, maybe I misread. Wait, the original square has vertices \((-2,2)\), \((2,2)\), \((2,-2)\), \((-2,-2)\). Let's check each transformation for whether the resulting vertices are the same set (maybe same points, maybe reordered).

Wait, no—maybe "same set of coordinate pairs" means the same four points (regardless of order). Let's re-examine:

Transformation 3: \((x, y) \to (-x, -y)\) (central symmetry)

Apply to each vertex:
- \((-2, 2) \to (2, -2)\)
- \((2, 2) \to (-2, -2)\)
- \((2, -2) \to (-2, 2)\)
- \((-2, -2) \to (2, 2)\)
These are the same four points (just permuted), so same set.

Transformation 4: \((x, y) \to (-y, x)\) (rotation 90 degrees counterclockwise)

Apply to each vertex:
- \((-2, 2) \to (-2, -2)\)
- \((2, 2) \to (-2, 2)\)
- \((2, -2) \to (2, 2)\)
- \((-2, -2) \to (2, -2)\)
Wait, no: \((x,y)\to(-y,x)\):
- For \((-2,2)\): \(-y = -2\), \(x = -2\) → \((-2, -2)\)
- For \((2,2)\): \(-y = -2\), \(x = 2\) → \((-2, 2)\)
- For \((2,-2)\): \(-y = 2\), \(x = 2\) → \((2, 2)\)
- For \((-2,-2)\): \(-y = 2\), \(x = -2\) → \((2, -2)\)
So the resulting vertices are \((-2,-2)\), \((-2,2)\), \((2,2)\), \((2,-2)\)—which is the same set as original (just rotated, but the points are the same four points, just ordered differently). Wait, no—the original points are \((-2,2)\), \((2,2)\), \((2,-2)\), \((-2,-2)\). The transformed points here are \((-2,-2)\), \((-2,2)\), \((2,2)\), \((2,-2)\)—same four points, just a different order. So that's the same set.

Wait, maybe I made a mistake with the second transformation: \((x,y)\to(x - 2, y - 2)\). Let's apply to each original vertex:

- \((-2,2)\): \(x-2 = -4\), \(y-2 = 0\) → \((-4, 0)\)
- \((2,2)\): \(x-2 = 0\), \(y-2 = 0\) → \((0, 0)\)
- \((2,-2)\): \(x-2 = 0\), \(y-2 = -4\) → \((0, -4)\)
- \((-2,-2)\): \(x-2 = -4\), \(y-2 = -4\) → \((-4, -4)\)
These are four new points (a square with side length 4, centered at \((-2, -2)\) shifted? No, this is a square, but the coordinate pairs are different (not the same set as original). So this transformation changes the coordinate pairs.

Wait, but let's check the first transformation: \((x,y)\to(-x, y)\) (reflection over y-axis). Apply to original vertices:

- \((-2,2)\to(2,2)\)
- \((2,2)\to(-2,2)\)
- \((2,-2)\to(-2,-2)\)
- \((-2,-2)\to(2,-2)\)
These are the same four points (just swapped left and right), so same set of coordinate pairs.

Transformation \((x,y)\to(-x, -y)\) (reflection over origin):

- \((-2,2)\to(2,-2)\)
- \((2,2)\to(-2,-2)\)
- \((2,-2)\to(-2,2)\)
- \((-2,-2)\to(2,2)\)
These are the same four points (permutation), so same set.

Transformation \((x,y)\to(-y, x)\) (rotation 90 degrees CCW):

As before, the points become \((-2,-2)\), \((-2,2)\), \((2,2)\), \((2,-2)\)—which is the same set (just rotated, so same four points).

But the transformation \((x,y)\to(x - 2, y - 2)\) translates the square, so the coordinate pairs are different (new points), so it does NOT produce a square with the same set of coordinate pairs. Wait, but let's check the options. The options are:

1. \((x,y)\to(-x, y)\) (reflection y-axis)
2. \((x,y)\to(x - 2, y - 2)\) (translation)
3. \((x,y)\to(-x, -y)\) (central symmetry)
4. \((x,y)\to(-y, x)\) (rotation 90 CCW)

Wait, the question is "Which transformation does NOT produce a square with the same set of coordinate pairs?"

So we need to see which transformation results in a square whose vertices are not the same four points (even in different order) as the original.

Original vertices: \(A(-2,2)\), \(B(2,2)\), \(C(2,-2)\), \(D(-2,-2)\).

- For \((x,y)\to(-x, y)\): New vertices are \(B(2,2)\), \(A(-2,2)\), \(D(-2,-2)\), \(C(2,-2)\)—same set (A,B,C,D → B,A,D,C; same four points).
- For \((x,y)\to(x - 2, y - 2)\): New vertices are \(A'(-4,0)\), \(B'(0,0)\), \(C'(0,-4)\), \(D'(-4,-4)\)—different set (new points), so this transformation does NOT produce a square with the same set of coordinate pairs.
- For \((x,y)\to(-x, -y)\): New vertices are \(C(2,-2)\), \(D(-2,-2)\), \(A(-2,2)\), \(B(2,2)\)—same set (permutation).
- For \((x,y)\to(-y, x)\): New vertices: \(D(-2,-2)\) → \((2, -2)\)? Wait no: \((x,y)\to(-y, x)\):

Wait, let's recalculate \((x,y)\to(-y, x)\) for each original vertex:

- \(A(-2,2)\): \(-y = -2\), \(x = -2\) → \((-2, -2)\) (which is D)
- \(B(2,2)\): \(-y = -2\), \(x = 2\) → \((-2, 2)\) (which is A)
- \(C(2,-2)\): \(-y = 2\), \(x = 2\) → \((2, 2)\) (which is B)
- \(D(-2,-2)\): \(-y = 2\), \(x = -2\) → \((2, -2)\) (which is C)

So the new vertices are D, A, B, C—same four points (A,B,C,D), just reordered. So same set.

Thus, the transformation \((x,y)\to(x - 2, y - 2)\) produces a square with different coordinate pairs (translated), so it does NOT produce a square with the same set of coordinate pairs.

Step2: Confirm each transformation

- \((x,y)\to(-x, y)\): Reflects over y-axis, same points.
- \((x,y)\to(-x, -y)\): Reflects over origin, same points (permutation).
- \((x,y)\to(-y, x)\): Rotates 90 CCW, same points (permutation).
- \((x,y)\to(x - 2, y - 2)\): Translates, new points.

Thus, the transformation \((x,y)\to(x - 2, y - 2)\) is the one that does NOT produce a square with the same set of coordinate pairs.

Answer:

\((x, y) \to (x - 2, y - 2)\) (the third option, assuming the options are ordered as: 1. \((x,y)\to(-x,y)\), 2. \((x,y)\to(x-2,y-2)\), 3. \((x,y)\to(-x,-y)\), 4. \((x,y)\to(-y,x)\) — but based on the image, the third option (from left) is \((x,y)\to(x - 2, y - 2)\))