Question

△sea is rotated 270° about the origin. draw the image of this rotation.

Explanation:

Step1: Identify Coordinates

First, find the coordinates of \( S \), \( E \), and \( A \). From the graph:
- \( S(-7, -2) \)
- \( E(-7, 2) \)
- \( A(-1, 2) \) (Wait, correction: Looking at the grid, \( A \) is at \( (-1, 2) \)? Wait no, original triangle: Wait, the blue triangle and the teal quadrilateral? Wait, \( \triangle SEA \): Let's recheck. \( S \) is at \( (-7, -2) \)? Wait, no, the teal points: \( E \) is at \( (-7, 2) \), \( S \) is at \( (-7, -2) \), and \( A \) is at \( (-1, 2) \)? Wait, no, the blue triangle: Wait, maybe I misread. Wait, the problem says \( \triangle SEA \). Let's look again. The teal points: \( E(-7, 2) \), \( S(-7, -2) \), and \( A(-1, 2) \)? Wait, no, the blue line from \( (0,3) \)? Wait, no, the graph has a blue triangle with vertex at \( (0,3) \), but \( \triangle SEA \) is teal? Wait, maybe the teal triangle is \( \triangle SEA \). So \( S(-7, -2) \), \( E(-7, 2) \), \( A(-1, 2) \). Wait, no, \( A \) is at \( (-1, 2) \)? Wait, the horizontal line from \( E(-7,2) \) to \( A \): \( A \) is at \( (-1, 2) \)? Yes, because the horizontal line is from \( x=-7 \) to \( x=-1 \), \( y=2 \). Then \( S \) is at \( (-7, -2) \), connected to \( E(-7,2) \) and \( A(-1,2) \).

Step2: Rotation Rule for \( 270^\circ \) about Origin

The rule for rotating a point \( (x, y) \) \( 270^\circ \) counterclockwise about the origin (or \( 90^\circ \) clockwise) is \( (x, y) \to (y, -x) \). Wait, no: Wait, \( 90^\circ \) counterclockwise: \( (x,y) \to (-y, x) \); \( 180^\circ \): \( (-x, -y) \); \( 270^\circ \) counterclockwise: \( (y, -x) \). Wait, let's confirm: Rotation of \( 270^\circ \) counterclockwise about origin: \( (x, y) \mapsto (y, -x) \).

So apply this to each vertex:

- For \( S(-7, -2) \):
- \( x = -7 \), \( y = -2 \)
- New coordinates: \( (y, -x) = (-2, 7) \)
- For \( E(-7, 2) \):
- \( x = -7 \), \( y = 2 \)
- New coordinates: \( (2, 7) \)
- For \( A(-1, 2) \):
- \( x = -1 \), \( y = 2 \)
- New coordinates: \( (2, 1) \)

Wait, wait, maybe I mixed up the rotation direction. Wait, \( 270^\circ \) counterclockwise is equivalent to \( 90^\circ \) clockwise. The rule for \( 90^\circ \) clockwise (or \( 270^\circ \) counterclockwise) is \( (x, y) \to (y, -x) \)? Wait, no, let's recall:

- \( 90^\circ \) counterclockwise: \( (x, y) \to (-y, x) \)
- \( 180^\circ \) counterclockwise: \( (x, y) \to (-x, -y) \)
- \( 270^\circ \) counterclockwise: \( (x, y) \to (y, -x) \) (Wait, no, actually, \( 270^\circ \) counterclockwise is \( (x, y) \to (y, -x) \)? Wait, let's test with a point. Take \( (1, 0) \), rotate \( 270^\circ \) counterclockwise: it should go to \( (0, 1) \)? Wait, no: \( 90^\circ \) counterclockwise: \( (1,0) \to (0,1) \); \( 180^\circ \): \( (-1, 0) \); \( 270^\circ \): \( (0, -1) \). Wait, so my previous rule was wrong. Let's derive it.

Rotation matrix for \( \theta \) counterclockwise: \( \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \). 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 \( (x, y) \):

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

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

Wait, but for \( (1, 0) \), \( x' = 0 \), \( y' = -1 \), which is \( (0, -1) \), correct (270° counterclockwise from (1,0) is (0,-1)).

Another test: \( (0, 1) \) rotated 270° counterclockwise: \( x' = 1 \), \( y' = 0 \), so (1, 0), correct.

So the rule is \( (x, y) \to (y, -x) \) for \( 270^\circ \) counterclockwise.

Now, let's re-express the coordinates of \( S \), \( E \), \( A \) correctly. Wait, maybe I misidentified the triangle. Let's look again: The teal triangle has vertices:

- \( S \): at \( (-7, -2) \) (since it's below the x-axis, \( y=-2 \), \( x=-7 \))
- \( E \): at \( (-7, 2) \) (above x-axis, \( y=2 \), \( x=-7 \))
- \( A \): at \( (-1, 2) \) (since the horizontal line from \( E(-7,2) \) to \( A \) is at \( y=2 \), \( x=-1 \))

So:

- \( S(-7, -2) \): apply rotation \( 270^\circ \) counterclockwise: \( (x, y) \to (y, -x) \) → \( (-2, 7) \) (since \( x=-7 \), \( y=-2 \); \( y' = -x = 7 \), \( x' = y = -2 \))
- \( E(-7, 2) \): \( (x, y) \to (y, -x) \) → \( (2, 7) \) ( \( x=-7 \), \( y=2 \); \( x' = 2 \), \( y' = 7 \))
- \( A(-1, 2) \): \( (x, y) \to (y, -x) \) → \( (2, 1) \) ( \( x=-1 \), \( y=2 \); \( x' = 2 \), \( y' = 1 \))

Wait, but let's check if the rotation is clockwise or counterclockwise. The problem says "rotated \( 270^\circ \) about the origin" – typically, unless specified, it's counterclockwise, but sometimes \( 270^\circ \) counterclockwise is same as \( 90^\circ \) clockwise. Let's confirm the rule for \( 90^\circ \) clockwise: \( (x, y) \to (y, -x) \), which is the same as \( 270^\circ \) counterclockwise. So that's correct.

Now, plot these new points:

- \( S'(-2, 7) \)
- \( E'(2, 7) \)
- \( A'(2, 1) \)

Wait, no, wait: \( S(-7, -2) \): \( x=-7 \), \( y=-2 \). So \( x' = y = -2 \), \( y' = -x = 7 \). So \( S'(-2, 7) \).

\( E(-7, 2) \): \( x=-7 \), \( y=2 \). \( x' = 2 \), \( y' = 7 \). So \( E'(2, 7) \).

\( A(-1, 2) \): \( x=-1 \), \( y=2 \). \( x' = 2 \), \( y' = 1 \). So \( A'(2, 1) \).

Now, connect these points to form the rotated triangle.

Wait, but maybe I made a mistake in the original coordinates. Let's re-express the original \( \triangle SEA \):

Looking at the graph, the teal triangle: \( S \) is at \( (-7, -2) \), \( E \) at \( (-7, 2) \), \( A \) at \( (-1, 2) \). Yes, because \( E \) and \( A \) are on \( y=2 \), \( S \) is directly below \( E \) at \( x=-7 \), \( y=-2 \), and connected to \( A \).

So after rotating \( 270^\circ \) counterclockwise, the new coordinates are:

- \( S'(-2, 7) \)
- \( E'(2, 7) \)
- \( A'(2, 1) \)

Now, plot these points:

- \( S'(-2, 7) \): \( x=-2 \), \( y=7 \)
- \( E'(2, 7) \): \( x=2 \), \( y=7 \)
- \( A'(2, 1) \): \( x=2 \), \( y=1 \)

Connecting these, the triangle will have a horizontal side from \( (-2,7) \) to \( (2,7) \), and a vertical side from \( (2,7) \) to \( (2,1) \), and a side from \( (2,1) \) to \( (-2,7) \).

Answer:

To draw the image of \( \triangle SEA \) rotated \( 270^\circ \) counterclockwise about the origin, follow these steps:

1. Identify Original Coordinates:

• \( S(-7, -2) \)
• \( E(-7, 2) \)
• \( A(-1, 2) \)

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

• \( S(-7, -2) \to S'(-2, 7) \) (since \( x=-7, y=-2 \); \( x' = y = -2 \), \( y' = -x = 7 \))
• \( E(-7, 2) \to E'(2, 7) \) (since \( x=-7, y=2 \); \( x' = y = 2 \), \( y' = -x = 7 \))
• \( A(-1, 2) \to A'(2, 1) \) (since \( x=-1, y=2 \); \( x' = y = 2 \), \( y' = -x = 1 \))

1. Plot and Connect New Points:

• Plot \( S'(-2, 7) \), \( E'(2, 7) \), and \( A'(2, 1) \) on the grid.
• Connect \( S' \) to \( E' \), \( E' \) to \( A' \), and \( A' \) to \( S' \) to form the rotated triangle.

(Note: If the rotation is clockwise \( 270^\circ \) (equivalent to counterclockwise \( 90^\circ \)), the rule is \( (x, y) \to (-y, x) \), but \( 270^\circ \) counterclockwise is standard here. Verify the direction if needed, but the above follows the rotation matrix for \( 270^\circ \) counterclockwise.)