Question

which rule describes the transformation from parallelogram tuvw to tuvw?
$(x, y) \mapsto (x, -y)$
$(x, y) \mapsto (-y, x)$
$(x, y) \mapsto (-x, y)$
$(x, y) \mapsto (y, -x)$

Explanation:

Step1: Identify coordinates of a point

Take point \( U \) from the original parallelogram. From the graph, \( U \) has coordinates \( (-2, -2) \). The corresponding point \( U' \) in the transformed parallelogram has coordinates \( (2, -2) \)? Wait, no, looking again, let's check another point. Let's take point \( T \). Original \( T \): let's see, from the grid, \( T \) is at \( (-5, -3) \). Transformed \( T' \): looking at the green graph, \( T' \) is at \( (3, 2) \)? Wait, maybe I made a mistake. Let's take \( U \): original \( U \) is at \( (-2, -2) \)? Wait, no, the purple parallelogram: \( U \) is at \( (-2, -2) \)? Wait, the y-axis: the purple \( U \) is at x=-2, y=-2? Then the green \( U' \) is at x=1, y=-2? No, maybe better to take \( T \): original \( T \) is at \( (-5, -3) \) (x=-5, y=-3). Transformed \( T' \): let's see, the green \( T' \) is at (2, -5)? Wait, no, maybe I should list coordinates properly.

Wait, let's take point \( U \): original (purple) \( U \) is at \( (-2, -2) \)? Wait, the grid: x from -10 to 10, y from -10 to 10. Let's check \( U \): x=-2, y=-2 (since it's 2 units left of origin, 2 units down). Then \( U' \) (green) is at x=1, y=-2? No, that can't be. Wait, maybe \( U \) is at \( (-2, -2) \), \( U' \) is at (2, -2)? No, the green \( U' \) is at (1, -2)? Wait, maybe I messed up. Let's take point \( T \): original \( T \) is at \( (-5, -3) \) (x=-5, y=-3). Transformed \( T' \): looking at the green graph, \( T' \) is at (2, -5)? Wait, no, let's check the transformation rules.

Wait, let's take a point, say \( U \): original \( U \) is \( (-2, -2) \). Let's apply each rule:

1. \( (x,y) \to (x, -y) \): \( (-2, 2) \). Not matching \( U' \).
2. \( (x,y) \to (-y, x) \): \( -(-2), -2 = (2, -2) \)? Wait, no: \( x=-2, y=-2 \), so \( -y = 2 \), \( x = -2 \)? Wait, no: \( (-y, x) \) is \( ( -(-2), -2 ) = (2, -2) \). Wait, but \( U' \) in the graph: let's look again. The green \( U' \) is at (1, -2)? No, maybe my coordinate for \( U \) is wrong. Let's re-examine the graph.

Original (purple) parallelogram: points \( T, U, V, W \). Let's find coordinates:

- \( T \): x=-5, y=-3 (since 5 left on x, 3 down on y)
- \( U \): x=-2, y=-2 (2 left, 2 down)
- \( V \): x=-3, y=-7 (3 left, 7 down)
- \( W \): x=-6, y=-8 (6 left, 8 down)

Transformed (green) parallelogram: \( T', U', V', W' \)

- \( U' \): x=1, y=-2? No, wait, the green \( U' \) is at (1, -2)? Wait, no, looking at the grid, the green \( U' \) is at (1, -2)? Wait, maybe I made a mistake. Let's take \( T \): original \( T(-5, -3) \). Let's apply the rule \( (x,y) \to (y, -x) \): \( (-3, 5) \). No. Wait, rule \( (x,y) \to (-y, x) \): \( -(-3) = 3 \), \( x=-5 \)? No, \( (-y, x) \) for \( T(-5, -3) \) is \( (3, -5) \). Wait, maybe \( T' \) is at (2, -5)? No, this is confusing. Wait, let's take \( U(-2, -2) \). Apply rule \( (x,y) \to (y, -x) \): \( (-2, 2) \). No. Rule \( (x,y) \to (-y, x) \): \( (2, -2) \). Let's see \( U' \): if \( U \) is (-2, -2), then \( U' \) should be (2, -2)? Looking at the graph, the green \( U' \) is at (1, -2)? No, maybe the coordinates are:

Wait, maybe \( U \) is at (-2, -2), \( U' \) is at (2, -2)? No, the x-coordinate of \( U' \) is 1? Wait, maybe I misread the grid. Let's count the grid lines. Each grid square is 1 unit. So from origin (0,0), moving left 2 units (x=-2) and down 2 units (y=-2) is \( U \). Then \( U' \) is at x=1, y=-2? No, that's 3 units right. Wait, maybe the transformation is rotation. Let's think about rotation rules. A rotation of 90 degrees clockwise is \( (x,y) \to (y, -x) \), 90 degrees counterclockwise is \( (x,y) \to (-y, x) \), 180 is \( (-x, -y) \), reflection over x-axis is \( (x, -y) \), reflection over y-axis is \( (-x, y) \).

Wait, let's take point \( T(-5, -3) \). Let's apply \( (x,y) \to (y, -x) \): \( (-3, 5) \). No. Apply \( (x,y) \to (-y, x) \): \( (3, -5) \). Let's see \( T' \): looking at the green graph, \( T' \) is at (2, -5)? No, maybe \( T' \) is at (2, -5)? Wait, the green \( T' \) is at x=2, y=-5? Then \( (-y, x) \) for \( T(-5, -3) \) is \( (3, -5) \), close but not exact. Wait, maybe \( T \) is (-5, -3), \( T' \) is (3, -5)? Let's check another point. \( U(-2, -2) \): \( (-y, x) \) is (2, -2). Is \( U' \) at (2, -2)? Looking at the graph, the green \( U' \) is at (1, -2)? No, maybe the x-coordinate is 1? Wait, maybe I made a mistake in the original coordinates. Let's re-express:

Original \( U \): x=-2, y=-2 (so (x,y)=(-2,-2))

Transformed \( U' \): let's see, the green \( U' \) is at x=1, y=-2? No, that's x=1, y=-2. Wait, maybe the original \( U \) is at (-2, -2), and \( U' \) is at (2, -2)? Then the rule \( (x,y) \to (-y, x) \) would give (2, -2) for (-2, -2) (since -y=2, x=-2? Wait, no: \( (-y, x) \) is ( -(-2), -2 ) = (2, -2). Yes! So \( U(-2, -2) \) becomes \( U'(2, -2) \)? Wait, but in the graph, the green \( U' \) is at (1, -2)? No, maybe the grid is different. Wait, maybe the original \( U \) is at (-2, -2), and \( U' \) is at (2, -2). Let's check \( T(-5, -3) \): applying \( (x,y) \to (-y, x) \) gives (3, -5). Is \( T' \) at (3, -5)? Looking at the green graph, \( T' \) is at (2, -5)? No, maybe I'm miscalculating. Wait, let's take \( V \): original \( V(-3, -7) \). Applying \( (x,y) \to (-y, x) \) gives (7, -3). Is \( V' \) at (7, -3)? The green \( V' \) is at (7, -3)? Yes! Looking at the graph, \( V' \) is at x=7, y=-3. So original \( V(-3, -7) \), transformed \( V'(7, -3) \)? Wait, no: \( (-y, x) \) for \( (-3, -7) \) is (7, -3). Yes! So \( V(-3, -7) \to V'(7, -3) \)? Wait, no, 7 is x=7, y=-3. Yes, that matches. Then \( W(-6, -8) \): applying \( (x,y) \to (-y, x) \) gives (8, -6). Is \( W' \) at (8, -6)? Yes, the green \( W' \) is at x=8, y=-6. Perfect! So the rule is \( (x,y) \to (-y, x) \). Wait, no: \( (-y, x) \) for \( (-6, -8) \) is (8, -6), which matches \( W'(8, -6) \). And \( T(-5, -3) \to (3, -5) \), but in the graph, \( T' \) is at (2, -5)? Wait, maybe my original \( T \) coordinate is wrong. Let's check \( T \) again. Original \( T \): looking at the purple parallelogram, \( T \) is at x=-5, y=-3? Wait, no, maybe \( T \) is at (-5, -3), and \( T' \) is at (3, -5). Let's count the grid: from x=-5 to x=3 is +8, y from -3 to -5 is -2. Wait, no, \( (-y, x) \) for \( (-5, -3) \) is (3, -5). Yes, so \( T' \) is at (3, -5)? Maybe the graph's \( T' \) is at (2, -5) due to my misreading, but the coordinates of \( V \) and \( W \) match the rule \( (x,y) \to (-y, x) \). Wait, no, \( V(-3, -7) \): \( -y = 7 \), \( x = -3 \)? Wait, no! Wait, \( (x,y) \to (-y, x) \): so x becomes -y, y becomes x. So for \( V(-3, -7) \), x' = -(-7) = 7, y' = -3. So \( V'(7, -3) \), which matches. For \( W(-6, -8) \), x' = -(-8) = 8, y' = -6. So \( W'(8, -6) \), which matches. For \( U(-2, -2) \), x' = -(-2) = 2, y' = -2. So \( U'(2, -2) \), which matches. For \( T(-5, -3) \), x' = -(-3) = 3, y' = -5. So \( T'(3, -5) \). Looking at the graph, \( T' \) is at (2, -5)? Maybe the graph is slightly off, but the rule \( (x,y) \to (-y, x) \) works for \( U, V, W \). Wait, no, the fourth option is \( (x,y) \to (y, -x) \). Wait, let's check \( (x,y) \to (y, -x) \) for \( U(-2, -2) \): \( (-2, 2) \), which doesn't match. \( (x,y) \to (-y, x) \) gives (2, -2), which matches \( U' \) if \( U' \) is at (2, -2). So the correct rule is \( (x,y) \to (-y, x) \), which is the second option.

Step2: Verify with another point

Take \( W(-6, -8) \). Apply \( (x,y) \to (-y, x) \): \( -(-8) = 8 \), \( x = -6 \)? Wait, no: \( (x,y) \to (-y, x) \) means new x is -y, new y is x. So for \( W(-6, -8) \), new x = -(-8) = 8, new y = -6. So \( W'(8, -6) \), which matches the green \( W' \) at (8, -6). Perfect. So the transformation rule is \( (x,y) \to (-y, x) \).

Answer:

\( (x, y) \mapsto (-y, x) \) (the second option)