Question

age under the given rotation about the origin.

1. trapezoid stuv with vertices s(-7, -1), t(-2, -3),

u(-2, -5), and v(-7, -7): 90° counterclockwise

s (__, __)
t (__, __)
u (__, __)
v (__, __)

Explanation:

Step1: Recall 90° CCW rotation rule

For a point \((x, y)\), 90° counterclockwise rotation about origin: \((x, y) \to (-y, x)\).

Step2: Rotate \( S(-7, -1) \)

Apply rule: \( x=-7, y=-1 \to (-(-1), -7)=(1, -7) \)? Wait, no: Wait, formula is \((x,y)\to(-y,x)\). So \( S(-7,-1) \): \( -y = -(-1)=1 \), \( x=-7 \)? Wait no, wait: 90° CCW: \((x,y) \mapsto (-y, x)\). So \( S(-7, -1) \): \( x=-7, y=-1 \). So new \( x = -y = -(-1) = 1 \), new \( y = x = -7 \)? Wait no, no: Wait, the correct formula for 90° counterclockwise rotation about the origin is \((x, y) ightarrow (-y, x)\). Let's verify with a simple point, like (1,0): 90° CCW is (0,1). Using formula: \( -y = 0 \), \( x = 1 \), so (0,1). Correct. Another: (0,1) 90° CCW is (-1,0). Formula: \( -y = -1 \), \( x=0 \), so (-1,0). Correct. So for \( S(-7, -1) \): \( x=-7 \), \( y=-1 \). New \( x = -y = -(-1) = 1 \), new \( y = x = -7 \)? Wait, no: Wait, the formula is \((x,y) \to (-y, x)\). So \( x \) becomes \( -y \), \( y \) becomes \( x \). So \( S(-7, -1) \): \( -y = -(-1) = 1 \), \( x = -7 \)? Wait, no, \( y \) component is \( x \), which is -7? Wait, no, original \( x \) is -7, so new \( y \) is \( x = -7 \)? Wait, no, let's do it again. Let's take \( (x, y) \). After 90° CCW, the new coordinates are \( (-y, x) \). So for \( S(-7, -1) \): \( x = -7 \), \( y = -1 \). So new \( x \): \( -y = -(-1) = 1 \). New \( y \): \( x = -7 \). So \( S' (1, -7) \)? Wait, no, wait, maybe I mixed up. Wait, no, let's check with (2,3). 90° CCW should be (-3,2). Using formula: \( -y = -3 \), \( x=2 \), so (-3,2). Correct. So yes, formula is \((x,y) \to (-y, x)\). So \( S(-7, -1) \): \( -y = -(-1) = 1 \), \( x = -7 \)? Wait, no, \( x \) in original is -7, so new \( y \) is \( x = -7 \). So \( S' (1, -7) \)? Wait, but let's check the graph. Wait, maybe I made a mistake. Wait, original points: S(-7,-1), T(-2,-3), U(-2,-5), V(-7,-7). Let's list each:

- \( S(-7, -1) \): apply \((x,y) \to (-y, x)\): \( -y = -(-1) = 1 \), \( x = -7 \)? Wait, no, \( x \) is -7, so new \( y \) is \( x = -7 \)? Wait, no, the formula is (x,y) becomes (-y, x). So x-coordinate of image is -y, y-coordinate is x. So for S(-7, -1): x=-7, y=-1. So image x: -(-1)=1, image y: -7. So S'(1, -7)? Wait, but let's check T(-2, -3): x=-2, y=-3. Image x: -(-3)=3, image y: -2. So T'(3, -2)? Wait, no, wait, maybe I have the formula reversed. Wait, 90° clockwise is (y, -x), 90° counterclockwise is (-y, x). Wait, let's confirm with (1,0): 90° CCW is (0,1). Using (-y, x): y=0, so -y=0, x=1, so (0,1). Correct. (0,1) 90° CCW: (-1, 0). -y=-1, x=0, so (-1,0). Correct. (1,1) 90° CCW: (-1,1). Correct. So formula is correct.

So let's process each point:

1. \( S(-7, -1) \):
- \( x = -7 \), \( y = -1 \)
- New \( x \): \( -y = -(-1) = 1 \)
- New \( y \): \( x = -7 \)
- So \( S'(1, -7) \)? Wait, no, wait, maybe I messed up the sign. Wait, no, let's take (2,3): 90° CCW is (-3,2). So x becomes -y (which is -3), y becomes x (which is 2). Correct. So for S(-7, -1): x=-7, y=-1. So -y = 1, x = -7. So S'(1, -7).

2. \( T(-2, -3) \):
- \( x = -2 \), \( y = -3 \)
- New \( x \): \( -y = -(-3) = 3 \)
- New \( y \): \( x = -2 \)
- So \( T'(3, -2) \)? Wait, no, wait, (2,3) becomes (-3,2). So x is -y, y is x. So T(-2,-3): x=-2, y=-3. So -y=3, x=-2. So T'(3, -2).

3. \( U(-2, -5) \):
- \( x = -2 \), \( y = -5 \)
- New \( x \): \( -y = -(-5) = 5 \)
- New \( y \): \( x = -2 \)
- So \( U'(5, -2) \)? Wait, no, (2,5) 90° CCW would be (-5,2). So U(-2,-5): x=-2, y=-5. -y=5, x=-2. So U'(5, -2).

4. \( V(-7, -7) \):
- \( x = -7 \), \( y = -7 \)
- New \( x \): \( -y = -(-7) = 7 \)
- New \( y \): \( x = -7 \)
- So \( V'(7, -7) \)? Wait, no, (7,7) 90° CCW is (-7,7). So V(-7,-7): x=-7, y=-7. -y=7, x=-7. So V'(7, -7).

Wait, but this seems odd. Wait, maybe the formula is (x,y) → (-y, x), but let's check the quadrant. Original points are in third quadrant (x negative, y negative). 90° CCW rotation would move them to fourth quadrant? Wait, no: third quadrant (x-, y-) rotated 90° CCW: the new x is -y (which is positive, since y is negative), new y is x (which is negative). So fourth quadrant (x+, y-). That makes sense.

Wait, but let's check with a point in third quadrant: (-1, -1). 90° CCW: (-(-1), -1) = (1, -1). Which is fourth quadrant. Correct. So the calculations seem right.

Wait, but maybe I made a mistake in the formula. Let me check another source: 90° counterclockwise rotation about origin: (x, y) → (-y, x). Yes, that's the standard formula.

So:

- \( S(-7, -1) \): \( (-(-1), -7) = (1, -7) \) → \( S'(1, -7) \)
- \( T(-2, -3) \): \( (-(-3), -2) = (3, -2) \) → \( T'(3, -2) \)
- \( U(-2, -5) \): \( (-(-5), -2) = (5, -2) \) → \( U'(5, -2) \)
- \( V(-7, -7) \): \( (-(-7), -7) = (7, -7) \) → \( V'(7, -7) \)

Wait, but let's plot these. Original S(-7,-1) is left and up a bit in third quadrant. Rotated 90° CCW: (1, -7) is right and down in fourth quadrant. T(-2,-3) rotated to (3, -2): right and up a bit in fourth. U(-2,-5) to (5, -2): right and up. V(-7,-7) to (7, -7): right and same y. That seems to form a trapezoid.

Alternatively, maybe I mixed up 90° CCW with 90° CW. Let's check 90° CW: (x,y)→(y, -x). For S(-7,-1): ( -1, 7 ). Which is second quadrant. But the problem says 90° counterclockwise, so CCW is correct as (-y, x).

So the coordinates should be:

\( S'(1, -7) \), \( T'(3, -2) \), \( U'(5, -2) \), \( V'(7, -7) \)

Wait, but let's check again. Wait, maybe I have the formula wrong. Wait, 90° counterclockwise rotation: the rotation matrix is \( \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \). So multiplying by vector \( \begin{pmatrix} x \\ y \end{pmatrix} \) gives \( \begin{pmatrix} -y \\ x \end{pmatrix} \). So yes, the formula is correct. So:

For \( S(-7, -1) \):

\( \begin{pmatrix} -(-1) \\ -7 \end{pmatrix} = \begin{pmatrix} 1 \\ -7 \end{pmatrix} \) → \( (1, -7) \)

For \( T(-2, -3) \):

\( \begin{pmatrix} -(-3) \\ -2 \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} \) → \( (3, -2) \)

For \( U(-2, -5) \):

\( \begin{pmatrix} -(-5) \\ -2 \end{pmatrix} = \begin{pmatrix} 5 \\ -2 \end{pmatrix} \) → \( (5, -2) \)

For \( V(-7, -7) \):

\( \begin{pmatrix} -(-7) \\ -7 \end{pmatrix} = \begin{pmatrix} 7 \\ -7 \end{pmatrix} \) → \( (7, -7) \)

Yes, that's correct.

Answer:

\( S'(1, -7) \), \( T'(3, -2) \), \( U'(5, -2) \), \( V'(7, -7) \)

So filling in the blanks:

\( S'(1, -7) \)

\( T'(3, -2) \)

\( U'(5, -2) \)

\( V'(7, -7) \)