Question

write the coordinates of the vertices after a rotation 90° counterclockwise around the origin.

Explanation:

Step1: Identify original coordinates

First, find the coordinates of each vertex. From the graph:
- \( S \): \( (-7, 1) \) (Wait, no, looking at the grid: Let's recheck. The x-axis: S is at x=-7? Wait, no, the grid lines: Let's see, T is at (-7,10)? Wait, no, the y-axis is up. Wait, the original figure: T, U, V, S. Let's check each:
- \( S \): x=-7, y=1? No, wait, the vertical lines: S is at x=-7, y=1? Wait, no, looking at the grid, S is at (-7, 1)? Wait, no, the yellow square: S is at (-7,1), T at (-7,10), U at (-2,10), V at (-2,1). Wait, no, the y-axis: the top is 10. So T: (-7,10), U: (-2,10), V: (-2,1), S: (-7,1).

Step2: Apply 90° counterclockwise rotation rule

The rule for 90° counterclockwise rotation around the origin is \( (x, y) \to (-y, x) \).

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

Apply the rule: \( x=-7, y=1 \to (-1, -7) \)? Wait, no: Wait, the rule is \( (x, y) \) becomes \( (-y, x) \). So \( x=-7, y=1 \): \( -y = -1 \), \( x = -7 \)? Wait, no, wait: Wait, 90° counterclockwise: the formula is \( (x, y) \mapsto (-y, x) \). Let's confirm with a point (a,b): rotating 90° counterclockwise around origin, the new coordinates are (-b, a).

So for \( S(-7, 1) \): \( x=-7, y=1 \). So new x: -y = -1, new y: x = -7. So \( S'(-1, -7) \)? Wait, no, wait, maybe I mixed up. Wait, let's take a point (2,3). Rotating 90° counterclockwise: (-3, 2). Yes, so (x,y) → (-y, x). So (2,3) → (-3,2). Correct.

So for \( T(-7, 10) \): \( x=-7, y=10 \). New x: -10, new y: -7. So \( T'(-10, -7) \)? Wait, no, wait: (-y, x) → (-10, -7). Wait, no: x=-7, y=10. So -y = -10, x = -7. So T'(-10, -7).

For \( U(-2, 10) \): \( x=-2, y=10 \). New x: -10, new y: -2. So \( U'(-10, -2) \).

For \( V(-2, 1) \): \( x=-2, y=1 \). New x: -1, new y: -2. So \( V'(-1, -2) \).

Wait, no, wait, maybe I made a mistake in original coordinates. Let's recheck the original coordinates. Let's look at the grid:

- The x-coordinate: S is at x=-7 (since it's 7 units left of origin), y=1 (since it's 1 unit up from x-axis). Wait, no, the y-axis: the bottom of S is at y=1 (since the grid lines: the x-axis is y=0, so S is at y=1. T is at y=10. So S(-7,1), T(-7,10), U(-2,10), V(-2,1).

Now apply rotation:

- \( S(-7, 1) \): \( (x,y) \to (-y, x) \) → (-1, -7)
- \( T(-7, 10) \): \( (-10, -7) \)
- \( U(-2, 10) \): \( (-10, -2) \)
- \( V(-2, 1) \): \( (-1, -2) \)

Wait, but that seems off. Wait, maybe the original coordinates are different. Wait, maybe the y-coordinate for S is 1? Wait, the grid: the x-axis is the horizontal line, y=0. So S is at (x,1), T at (x,10), U at (-2,10), V at (-2,1). So x for S and T is -7, x for U and V is -2. So S(-7,1), T(-7,10), U(-2,10), V(-2,1).

Wait, let's recheck the rotation rule. Maybe I had the rule reversed. Wait, 90° clockwise is (y, -x), 90° counterclockwise is (-y, x). Yes. So (x,y) → (-y, x) for 90° counterclockwise.

So let's re-calculate:

- \( S(-7, 1) \): \( x=-7, y=1 \). New x: -y = -1, new y: x = -7. So \( S'(-1, -7) \)
- \( T(-7, 10) \): \( x=-7, y=10 \). New x: -10, new y: -7. So \( T'(-10, -7) \)
- \( U(-2, 10) \): \( x=-2, y=10 \). New x: -10, new y: -2. So \( U'(-10, -2) \)
- \( V(-2, 1) \): \( x=-2, y=1 \). New x: -1, new y: -2. So \( V'(-1, -2) \)

Wait, but that seems to move the points to the third quadrant. Let's check with a simple point: (1,0). Rotating 90° counterclockwise: (0,1)? No, wait, (1,0) rotated 90° counterclockwise: (0,1)? Wait, no, (1,0) is on x-axis. Rotating 90° counterclockwise around origin: (0,1). Using the rule: (x,y)=(1,0) → (-y, x)= (0,1). Correct. (0,1) rotated 90° counterclockwise: (-1,0). Correct. (-1,0) rotated 90° counterclockwise: (0,-1). Correct. (0,-1) rotated 90° counterclockwise: (1,0). Correct. So the rule holds.

So applying to the points:

Original coordinates:
- \( S(-7, 1) \)
- \( T(-7, 10) \)
- \( U(-2, 10) \)
- \( V(-2, 1) \)

After 90° counterclockwise rotation:

- \( S'(-1, -7) \)
- \( T'(-10, -7) \)
- \( U'(-10, -2) \)
- \( V'(-1, -2) \)

Wait, but maybe I made a mistake in the original coordinates. Let's re-examine the graph. The x-axis: the leftmost is -10, rightmost 10. The y-axis: top 10, bottom -10. The yellow square: S is at x=-7, y=1 (since it's on the line y=1, x=-7). T is at x=-7, y=10 (on y=10, x=-7). U is at x=-2, y=10 (x=-2, y=10). V is at x=-2, y=1 (x=-2, y=1). So that's correct.

So applying the rotation rule:

For each point (x, y), new coordinates are (-y, x).

So:

- S(-7, 1): (-1, -7)
- T(-7, 10): (-10, -7)
- U(-2, 10): (-10, -2)
- V(-2, 1): (-1, -2)

Answer:

The coordinates after rotation are:

• \( S'(-1, -7) \)
• \( T'(-10, -7) \)
• \( U'(-10, -2) \)
• \( V'(-1, -2) \)