Question

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

Explanation:

Step1: Identify original coordinates

First, find the original coordinates of each vertex. From the graph:
- \( D(5, 7) \) (wait, no, looking at the grid: D is at (5,7)? Wait, x-axis: D is at x=5, y=7? Wait, no, the grid: let's check again. The blue square: D is at (5,7)? Wait, E is at (5,10)? Wait, no, the y-axis: E is at y=10? Wait, the grid lines: each square is 1 unit. Let's re-express:

Looking at the graph:
- \( D \): x=5, y=7 (since it's on x=5, y=7)
- \( E \): x=5, y=10 (x=5, y=10)
- \( F \): x=8, y=10 (x=8, y=10)
- \( G \): x=8, y=7 (x=8, y=7)

Wait, maybe I misread. Let's check the y-axis: the top of E is at y=10? Wait, the y-axis has 10 at the top. So:

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

Step2: Rotation rule for 270° counterclockwise

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, let's confirm:

Rotation 270° counterclockwise is equivalent to rotation 90° clockwise. The rule for 90° clockwise (which is same as 270° counterclockwise) is \( (x, y) ightarrow (y, -x) \)? Wait, no: let's recall the rotation matrices.

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 \( (x, y) \), we get \( (y, -x) \). Wait, let's test with a point. For example, (1,0) rotated 270° counterclockwise: should be (0,1)? Wait, no: (1,0) rotated 90° counterclockwise is (0,1), 180° is (-1,0), 270° is (0,-1). Wait, maybe I mixed up. Wait, 270° counterclockwise: let's think of rotating (x,y) 270° counterclockwise. Let's take (1,0): 270° counterclockwise around origin: the point moves to (0,-1). Using the matrix: \( \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ -1 \end{pmatrix} \), which is (0,-1). So the rule is \( (x, y) ightarrow (y, -x) \)? Wait, (1,0) becomes (0, -1), which is (y, -x) since y=0, -x=-1. Yes. So the rule is \( (x, y) ightarrow (y, -x) \).

Wait, another example: (0,1) rotated 270° counterclockwise: should be (1, 0). Using the rule: (y, -x) = (1, 0), which is correct. (1,1) rotated 270° counterclockwise: (1, -1)? Wait, no: (1,1) rotated 90° counterclockwise is (-1,1), 180° is (-1,-1), 270° is (1,-1). Using the rule: (y, -x) = (1, -1), correct. So the rule is \( (x, y) ightarrow (y, -x) \).

Step3: Apply rotation to each point

Now apply the rule \( (x, y) ightarrow (y, -x) \) to each vertex:

- For \( D(5, 7) \):
\( x = 5 \), \( y = 7 \)
New coordinates: \( (7, -5) \) → \( D'(7, -5) \)

- For \( E(5, 10) \):
\( x = 5 \), \( y = 10 \)
New coordinates: \( (10, -5) \) → \( E'(10, -5) \)

- For \( F(8, 10) \):
\( x = 8 \), \( y = 10 \)
New coordinates: \( (10, -8) \) → \( F'(10, -8) \)

- For \( G(8, 7) \):
\( x = 8 \), \( y = 7 \)
New coordinates: \( (7, -8) \) → \( G'(7, -8) \)

Wait, but let's verify again. Wait, maybe I made a mistake in the original coordinates. Let's recheck the original points:

Looking at the graph:

- \( D \): x=5, y=7? Wait, the y-axis: D is at y=7? Wait, the blue square: D is connected to E (above) and G (right). So D is (5,7), E is (5,10), F is (8,10), G is (8,7). Yes. So:

- \( D(5,7) \): rotate 270° CCW: (7, -5)
- \( E(5,10) \): (10, -5)
- \( F(8,10) \): (10, -8)
- \( G(8,7) \): (7, -8)

Wait, but let's check with another approach. Rotation 270° counterclockwise is the same as rotation 90° clockwise. The rule for 90° clockwise is \( (x, y) ightarrow (y, -x) \), which matches. So that's correct.

Answer:

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

(Note: If the original coordinates were misread, let's recheck. Wait, maybe D is (5,7)? Wait, the x-axis: D is at x=5 (since between 4 and 6, so 5), y=7 (between 6 and 8, so 7). E is at x=5, y=10 (top of the square). F is at x=8, y=10. G is at x=8, y=7. So the rotation is correct.)