Question

graph the image of rectangle stuv after a rotation 270° counterclockwise around the origin.

Explanation:

First, we need to find the coordinates of the vertices of rectangle \( STUV \). From the graph:
- \( S(-6, -9) \) (assuming the bottom vertex S is at y=-9, since it's a rectangle and V is at (-6, -1), wait, no, looking at the grid, V is at (-6, -1)? Wait, no, the grid: V is at (-6, -1)? Wait, the y-axis: the orange points. Let's recheck:

Looking at the graph, the coordinates:
- \( V(-6, -1) \)? Wait, no, the vertical lines: V is at (-6, -1)? Wait, no, the y-coordinate: the orange segment from V to S: V is at (-6, -1), S is at (-6, -9)? Wait, no, the grid lines: each grid is 1 unit. Let's see:

Point \( V \): x=-6, y=-1? Wait, no, the point V is at (-6, -1)? Wait, the U is at (-3, -1)? Wait, no, the original rectangle: V is at (-6, -1), U is at (-3, -1), T is at (-3, -9), S is at (-6, -9)? Wait, no, the graph shows:

Wait, the orange points: V is at (-6, -1) (since it's on the horizontal line y=-1, x=-6), U is at (-3, -1) (x=-3, y=-1), T is at (-3, -9) (x=-3, y=-9), S is at (-6, -9) (x=-6, y=-9). Yes, that makes a rectangle.

Now, the rule for a \( 270^\circ \) counterclockwise rotation around the origin is \( (x, y) ightarrow (y, -x) \).

Step 1: Find coordinates of each vertex

- \( S(-6, -9) \)
- \( T(-3, -9) \)
- \( U(-3, -1) \)
- \( V(-6, -1) \)

Step 2: Apply rotation rule \( (x, y)

ightarrow (y, -x) \) to each vertex
- For \( S(-6, -9) \):
New x = -9, New y = -(-6) = 6. So \( S'(-9, 6) \)? Wait, no, wait the rotation rule: \( 270^\circ \) counterclockwise is equivalent to \( 90^\circ \) clockwise, and the rule is \( (x, y) ightarrow (y, -x) \). Wait, let's confirm:

The rotation rules:
- \( 90^\circ \) counterclockwise: \( (x, y) ightarrow (-y, x) \)
- \( 180^\circ \) counterclockwise: \( (x, y) ightarrow (-x, -y) \)
- \( 270^\circ \) counterclockwise: \( (x, y) ightarrow (y, -x) \)

Yes, so:

- \( S(-6, -9) \): \( (x, y) = (-6, -9) \), so new \( (y, -x) = (-9, -(-6)) = (-9, 6) \)? Wait, no, \( -x \) when x=-6 is -(-6)=6. So \( S'(-9, 6) \)?

Wait, no, let's take a point (a, b). After \( 270^\circ \) counterclockwise rotation, the new coordinates are (b, -a). So for (x, y) = (a, b), new (b, -a).

So:

- \( S(-6, -9) \): a=-6, b=-9. New (b, -a) = (-9, -(-6)) = (-9, 6)
- \( T(-3, -9) \): a=-3, b=-9. New (b, -a) = (-9, -(-3)) = (-9, 3)
- \( U(-3, -1) \): a=-3, b=-1. New (b, -a) = (-1, -(-3)) = (-1, 3)
- \( V(-6, -1) \): a=-6, b=-1. New (b, -a) = (-1, -(-6)) = (-1, 6)

Wait, but let's check with a simple point. For example, a point (1, 0) rotated \( 270^\circ \) counterclockwise: it should go to (0, -1)? Wait, no, \( 270^\circ \) counterclockwise from (1,0) is (0, -1)? Wait, no, \( 90^\circ \) counterclockwise: (0,1); \( 180^\circ \): (-1,0); \( 270^\circ \): (0, -1). Using the rule (x,y)→(y, -x): (1,0)→(0, -1). Correct. So the rule is correct.

Now, let's re-express the original coordinates correctly. Wait, maybe I misread the coordinates. Let's look again:

The rectangle: V is at (-6, -1), U is at (-3, -1), T is at (-3, -9), S is at (-6, -9). So:

- \( V(-6, -1) \)
- \( U(-3, -1) \)
- \( T(-3, -9) \)
- \( S(-6, -9) \)

Now apply the rotation:

- \( S(-6, -9) \): (x,y)=(-6,-9) → (y, -x)=(-9, -(-6))=(-9, 6)
- \( T(-3, -9) \): (x,y)=(-3,-9) → (y, -x)=(-9, -(-3))=(-9, 3)
- \( U(-3, -1) \): (x,y)=(-3,-1) → (y, -x)=(-1, -(-3))=(-1, 3)
- \( V(-6, -1) \): (x,y)=(-6,-1) → (y, -x)=(-1, -(-6))=(-1, 6)

Now, we need to plot these new points:

- \( S'(-9, 6) \)
- \( T'(-9, 3) \) Wait, no, wait T was (-3, -9), so (y, -x) is (-9, 3). Wait, no, T is (-3, -9), so x=-3, y=-9. So (y, -x) = (-9, 3). Correct.

Wait, no, U is (-3, -1): x=-3, y=-1. So (y, -x) = (-1, 3). Correct.

V is (-6, -1): x=-6, y=-1. So (y, -x) = (-1, 6). Correct.

S is (-6, -9): x=-6, y=-9. So (y, -x) = (-9, 6). Correct.

Now, let's check the rectangle. The original rectangle has length 3 (from x=-6 to x=-3) and height 8 (from y=-1 to y=-9). After rotation, the length and height should be preserved, but rotated.

Wait, maybe I made a mistake in the original coordinates. Let's check the grid again. The vertical lines: from y=-1 to y=-9 is 8 units (since -1 - (-9) = 8). The horizontal lines: from x=-6 to x=-3 is 3 units (since -3 - (-6) = 3).

After rotation \( 270^\circ \) counterclockwise, the horizontal length becomes vertical and vertical height becomes horizontal. Wait, maybe the original coordinates are different. Wait, maybe the y-coordinate of V is -1? Wait, the grid: the y-axis has -1, -2, ..., -10. The orange points: V is at (-6, -1), U at (-3, -1), T at (-3, -9), S at (-6, -9). Yes, that's correct.

So after rotation, the new points are:

- \( S'(-9, 6) \)
- \( T'(-9, 3) \)? No, wait T is (-3, -9), so (y, -x) is (-9, 3). Wait, no, T is (-3, -9), so x=-3, y=-9. So (y, -x) = (-9, 3). So T' is (-9, 3)? Wait, no, that can't be. Wait, maybe the rotation rule is \( 270^\circ \) clockwise, which is \( (x, y) ightarrow (y, -x) \) as well? Wait, no, \( 270^\circ \) counterclockwise is the same as \( 90^\circ \) clockwise, and the rule is \( (x, y) ightarrow (y, -x) \).

Wait, let's take a simple point (2, 3). Rotate \( 270^\circ \) counterclockwise: new point (3, -2). Let's check with a graph: (2,3) is in the first quadrant. Rotating \( 270^\circ \) counterclockwise (or \( 90^\circ \) clockwise) would move it to the fourth quadrant? Wait, no, (3, -2) is in the fourth quadrant. Correct.

Another point: (-2, 3). Rotate \( 270^\circ \) counterclockwise: (3, 2). Let's see: (-2,3) is in the second quadrant. Rotating \( 270^\circ \) counterclockwise: (3, 2) is in the first quadrant. Correct, because \( 270^\circ \) counterclockwise from second quadrant: 90° (first), 180° (third), 270° (fourth? No, wait, 270° counterclockwise from (-2,3):

The rotation matrix for \( 270^\circ \) counterclockwise is \( \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \). So applying to vector \( \begin{pmatrix} x \\ y \end{pmatrix} \), we get \( \begin{pmatrix} y \\ -x \end{pmatrix} \). So for (-2, 3), we get \( \begin{pmatrix} 3 \\ 2 \end{pmatrix} \), which is (3, 2). Correct.

So the rule is correct.

Now, back to the original points:

- \( V(-6, -1) \): \( \begin{pmatrix} -6 \\ -1 \end{pmatrix} \) → \( \begin{pmatrix} -1 \\ 6 \end{pmatrix} \) (since -x = -(-6) = 6)
- \( U(-3, -1) \): \( \begin{pmatrix} -3 \\ -1 \end{pmatrix} \) → \( \begin{pmatrix} -1 \\ 3 \end{pmatrix} \) (since -x = -(-3) = 3)
- \( T(-3, -9) \): \( \begin{pmatrix} -3 \\ -9 \end{pmatrix} \) → \( \begin{pmatrix} -9 \\ 3 \end{pmatrix} \) (since -x = -(-3) = 3)
- \( S(-6, -9) \): \( \begin{pmatrix} -6 \\ -9 \end{pmatrix} \) → \( \begin{pmatrix} -9 \\ 6 \end{pmatrix} \) (since -x = -(-6) = 6)

Now, plotting these points:

- \( V'(-1, 6) \)
- \( U'(-1, 3) \)
- \( T'(-9, 3) \)
- \( S'(-9, 6) \)

Wait, that makes a rectangle with length 8 (from y=3 to y=6 is 3 units? No, wait the original height was 8 (from y=-1 to y=-9 is 8 units). After rotation, the length should be 8 and width 3. Wait, the new points:

- \( S'(-9, 6) \)
- \( T'(-9, 3) \): vertical distance 3 (6-3=3)? No, that's not 8. So I must have messed up the original coordinates.

Ah! Here's the mistake: the original y-coordinates. Let's re-examine the graph. The orange points: V is at (-6, -1)? No, looking at the grid, the y-axis has -1, -2, ..., -10. The point V is at (-6, -1)? Wait, the vertical segment from V to S: from y=-1 to y=-9? That's 8 units (since -1 - (-9) = 8). The horizontal segment from V to U: from x=-6 to x=-3, which is 3 units (since -3 - (-6) = 3). So that's correct.

But after rotation, the vertical segment (height 8) becomes horizontal, and horizontal segment (length 3) becomes vertical. Wait, no, \( 270^\circ \) counterclockwise rotation: a horizontal line (along x-axis) becomes vertical (along y-axis), and vertical line (along y-axis) becomes horizontal (along x-axis).

Wait, the original rectangle has a horizontal side from V(-6, -1) to U(-3, -1) (length 3, horizontal) and vertical side from U(-3, -1) to T(-3, -9) (length 8, vertical). After \( 270^\circ \) counterclockwise rotation, the horizontal side (length 3) becomes vertical (length 3), and the vertical side (length 8) becomes horizontal (length 8).

So applying the rotation rule to U(-3, -1) and T(-3, -9):

- U(-3, -1): (x,y)=(-3,-1) → (y, -x)=(-1, 3)
- T(-3, -9): (x,y)=(-3,-9) → (y, -x)=(-9, 3)

So the segment from U'(-1, 3) to T'(-9, 3) is horizontal, length 8 (from x=-9 to x=-1 is 8 units: -1 - (-9) = 8). Correct, because the original vertical segment was length 8 (from y=-1 to y=-9 is 8 units: -1 - (-9) = 8).

The segment from V(-6, -1) to S(-6, -9): vertical, length 8. After rotation:

- V(-6, -1): (x,y)=(-6,-1) → (y, -x)=(-1, 6)
- S(-6, -9): (x,y)=(-6,-9) → (y, -x)=(-9, 6)

So the segment from V'(-1, 6) to S'(-9, 6) is horizontal, length 8 (from x=-9 to x=-1 is 8 units: -1 - (-9) = 8). Correct.

The segment from V'(-1, 6) to U'(-1, 3) is vertical, length 3 (from y=3 to y=6 is 3 units: 6 - 3 = 3). Correct, because the original horizontal segment was length 3 (from x=-6 to x=-3 is 3 units: -3 - (-6) = 3).

So the new rectangle has vertices at:

- \( V'(-1, 6) \)
- \( U'(-1, 3) \)
- \( T'(-9, 3) \)
- \( S'(-9, 6) \)

Now, we can plot these points on the grid.

Answer:

The image of rectangle \( STUV \) after a \( 270^\circ \) counterclockwise rotation around the origin has vertices at \( S'(-9, 6) \), \( T'(-9, 3) \), \( U'(-1, 3) \), and \( V'(-1, 6) \). (To graph, plot these points and connect them in order.)