Question

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

Explanation:

Step1: Identify coordinates

First, find the coordinates of rectangle \( STUV \). From the graph:

• \( S(2, -9) \)
• \( T(8, -9) \)
• \( U(8, -7) \)
• \( V(2, -7) \)

Step2: Apply 270° counterclockwise rotation rule

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

For \( S(2, -9) \):

New coordinates: \( (-9, -2) \) (Wait, no: Wait, the rule is \( (x,y) \to (y, -x) \). So \( x = 2 \), \( y = -9 \). So new \( x = y = -9 \), new \( y = -x = -2 \). So \( S'( -9, -2) \)? Wait, no, wait: Wait, 270° counterclockwise is equivalent to 90° clockwise. The correct rule for 270° counterclockwise around origin is \( (x, y) \mapsto (y, -x) \). Let's recheck:

Wait, standard rotation rules:

• 90° counterclockwise: \( (x, y) \to (-y, x) \)
• 180° counterclockwise: \( (x, y) \to (-x, -y) \)
• 270° counterclockwise: \( (x, y) \to (y, -x) \)

So for \( S(2, -9) \):
\( x = 2 \), \( y = -9 \)
New \( x = y = -9 \)
New \( y = -x = -2 \)
So \( S'(-9, -2) \)? Wait, no, that seems off. Wait, maybe I mixed up. Wait, 270° counterclockwise is the same as 90° clockwise. The rule for 90° clockwise is \( (x, y) \to (y, -x) \). Wait, let's take a point (1,0). Rotate 270° counterclockwise: it should go to (0,1)? Wait, no. Wait, (1,0) rotated 90° counterclockwise is (0,1), 180° is (-1,0), 270° is (0,-1). Wait, using the rule \( (x,y) \to (y, -x) \): (1,0) becomes (0, -1), which is correct. So the rule is correct.

Now for each point:

• \( S(2, -9) \): \( (x,y) = (2, -9) \to (y, -x) = (-9, -2) \) → \( S'(-9, -2) \)
• \( T(8, -9) \): \( (8, -9) \to (-9, -8) \) → \( T'(-9, -8) \)
• \( U(8, -7) \): \( (8, -7) \to (-7, -8) \) → \( U'(-7, -8) \)
• \( V(2, -7) \): \( (2, -7) \to (-7, -2) \) → \( V'(-7, -2) \)

Wait, that seems to place the points in the third quadrant. Wait, maybe I made a mistake. Wait, let's check the original coordinates again. From the graph, the rectangle is at the bottom, with S at (2, -9)? Wait, the y-axis: the bottom is -10, so S is at (2, -9)? Wait, the grid lines: each grid is 1 unit. So S is at (2, -9), T at (8, -9), U at (8, -7), V at (2, -7). So that's a rectangle with length 6 (from x=2 to x=8) and height 2 (from y=-9 to y=-7).

Now, rotating 270° counterclockwise around the origin. Let's use the correct rotation matrix. The rotation matrix for \( \theta \) counterclockwise is \(

$$\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$

\). For \( \theta = 270^\circ \), \( \cos 270^\circ = 0 \), \( \sin 270^\circ = -1 \). So the matrix is \(

$$\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$

\). So applying to a point \( (x, y) \):

\( x' = 0 \cdot x + 1 \cdot y = y \)

\( y' = -1 \cdot x + 0 \cdot y = -x \)

So the rule \( (x, y) \to (y, -x) \) is correct.

So:

• \( S(2, -9) \): \( x' = -9 \), \( y' = -2 \) → \( S'(-9, -2) \)
• \( T(8, -9) \): \( x' = -9 \), \( y' = -8 \) → \( T'(-9, -8) \)
• \( U(8, -7) \): \( x' = -7 \), \( y' = -8 \) → \( U'(-7, -8) \)
• \( V(2, -7) \): \( x' = -7 \), \( y' = -2 \) → \( V'(-7, -2) \)

Now, plot these points:

• \( S'(-9, -2) \)
• \( T'(-9, -8) \)
• \( U'(-7, -8) \)
• \( V'(-7, -2) \)

Connecting these points will give the rotated rectangle.

Answer:

To graph the image of rectangle \( STUV \) after a \( 270^\circ \) counterclockwise rotation around the origin, follow these steps:

1. Identify Original Coordinates:

• \( S(2, -9) \), \( T(8, -9) \), \( U(8, -7) \), \( V(2, -7) \).

1. Apply Rotation Rule (\( (x, y) \to (y, -x) \) for \( 270^\circ \) counterclockwise):

• \( S(2, -9) \to S'(-9, -2) \)
• \( T(8, -9) \to T'(-9, -8) \)
• \( U(8, -7) \to U'(-7, -8) \)
• \( V(2, -7) \to V'(-7, -2) \)

1. Plot and Connect Points:

Plot \( S'(-9, -2) \), \( T'(-9, -8) \), \( U'(-7, -8) \), \( V'(-7, -2) \), then connect them to form the rotated rectangle.

(Note: The final graph will have vertices at \( (-9, -2) \), \( (-9, -8) \), \( (-7, -8) \), and \( (-7, -2) \).)