Question

write the coordinates of the vertices after a rotation 270° counterclockwise around the origin.
d(\square, \square)
e(\square, \square)
f(\square, \square)
g(\square, \square)

Explanation:

Step1: Find original coordinates

First, identify the original coordinates of each vertex from the graph:

• \( D(5, 7) \) (Wait, no, looking at the grid: Let's recheck. The x-axis and y-axis: For point D, x=5? Wait, the grid lines: E is at (5,10)? Wait no, the graph: E is at (5,10)? Wait no, the y-axis has 10 at the top. Wait, the points: D is at (5,7)? Wait no, let's look again. The square: D, E, F, G. Let's see: D is (5,7)? Wait, no, the y-coordinate for D: the horizontal line for D is at y=7? Wait, the grid: each square is 1 unit. So E is at (5,10)? Wait, no, the y-axis: from 0 to 10, so E is (5,10), F is (8,10), G is (8,7), D is (5,7). Yes, that makes sense. So original coordinates:
• \( D(5, 7) \)
• \( E(5, 10) \)
• \( F(8, 10) \)
• \( G(8, 7) \)

Step2: Recall 270° counterclockwise rotation rule

The rule for rotating a point \( (x, y) \) 270° counterclockwise around the origin is \( (x, y)
ightarrow (y, -x) \). Wait, no: Wait, 90° counterclockwise is \( (-y, x) \), 180° is \( (-x, -y) \), 270° counterclockwise is \( (y, -x) \)? Wait, no, let's confirm: The rotation matrix for 270° counterclockwise is \(

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

\), so applying to \( (x, y) \), we get \( (y, -x) \). Wait, no: Wait, 90° counterclockwise: \( (x, y)
ightarrow (-y, x) \)
180° counterclockwise: \( (x, y)
ightarrow (-x, -y) \)
270° counterclockwise: \( (x, y)
ightarrow (y, -x) \)? Wait, no, let's test with a point. Take (1,0), rotate 270° counterclockwise: it should go to (0,1)? Wait, no. Wait, 270° counterclockwise is the same as 90° clockwise. The rule for 90° clockwise is \( (x, y)
ightarrow (y, -x) \)? Wait, no: 90° clockwise: \( (x, y)
ightarrow (y, -x) \)? Wait, let's take (1,0): 90° clockwise is (0, -1)? No, that's not right. Wait, I think I mixed up. Let's use the correct rotation rules:

• 90° counterclockwise: \( (x, y)

ightarrow (-y, x) \)

• 180° counterclockwise: \( (x, y)

ightarrow (-x, -y) \)

• 270° counterclockwise: \( (x, y)

ightarrow (y, -x) \)? Wait, no, let's use a point (2,3). Rotate 270° counterclockwise: the result should be (3, -2). Let's check with the rotation matrix. The rotation matrix for θ degrees counterclockwise is \(

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

\). For θ=270°, \( \cos270°=0 \), \( \sin270°=-1 \). So the matrix is \(

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

\). So applying to (x,y): \( x' = 0*x + 1*y = y \), \( y' = -1*x + 0*y = -x \). So yes, 270° counterclockwise rotation rule is \( (x, y)
ightarrow (y, -x) \).

Step3: Apply the rule to each point

• For \( D(5, 7) \):

\( x' = 7 \), \( y' = -5 \), so \( D'(7, -5) \)

• For \( E(5, 10) \):

\( x' = 10 \), \( y' = -5 \), so \( E'(10, -5) \)

• For \( F(8, 10) \):

\( x' = 10 \), \( y' = -8 \), so \( F'(10, -8) \)

• For \( G(8, 7) \):

\( x' = 7 \), \( y' = -8 \), so \( G'(7, -8) \)

Wait, let's verify with a simpler point. Take (0,1), rotate 270° counterclockwise: using the rule, (1, 0). Which is correct, because (0,1) rotated 270° counterclockwise (or 90° clockwise) is (1,0). Yes, that works. Another test: (1,1) rotated 270° counterclockwise: (1, -1). Let's see: using the matrix, \( x' = 1 \), \( y' = -1 \). If we rotate (1,1) 270° counterclockwise, it should end up at (1, -1). Let's visualize: (1,1) is in the first quadrant. Rotating 270° counterclockwise (which is three 90° turns counterclockwise) would move it to (1, -1), which is in the fourth quadrant. That makes sense. So the rule seems correct.

So applying to the points:

• D(5,7) → (7, -5)
• E(5,10) → (10, -5)
• F(8,10) →…

Answer:

\( D'(7, -5) \), \( E'(10, -5) \), \( F'(10, -8) \), \( G'(7, -8) \)