Question

point ( a ) is the image of point ( a ) under a rotation about the origin, ( (0,0) ).
determine the angles of rotation.
choose all answers that apply:
a ( 90^circ ) clockwise
b ( 90^circ ) counterclockwise
c ( 180^circ )
d ( 270^circ ) clockwise
e ( 270^circ ) counterclockwise

Explanation:

Step1: Identify coordinates of A and A'

Point \( A \) is at \( (-1, 5) \) (wait, no, looking at the grid: A is at (-1,5)? Wait, no, the grid: x-axis from -7 to 7, y-axis from -7 to 7. A is at (-1,5)? Wait, no, the blue dot A: let's check the grid. The x-coordinate: between -2 and 0, so x=-1? Wait, no, the grid lines: each square is 1 unit. So A is at (-1, 5)? Wait, no, the pink dot A' is at (5,1)? Wait, no, A' is at (5,1)? Wait, no, looking at the graph: A is at (-1, 5)? Wait, no, the y-axis is up, x-axis right. A is at (-1, 5)? Wait, no, the blue dot A: x=-1, y=5? Wait, A' is at (5,1)? Wait, no, let's check again. Wait, A is at (-1, 5)? Wait, no, the grid: A is at ( -1, 5 )? Wait, A' is at (5, 1)? Wait, no, maybe I misread. Wait, A is at (-1, 5)? Wait, no, the x-coordinate: from the origin (0,0), moving left 1 unit (x=-1), up 5 units (y=5). A' is at (5,1): right 5 units, up 1 unit? Wait, no, the pink dot A' is at (5,1)? Wait, no, looking at the graph, A' is at (5,1)? Wait, no, maybe A is at (-1, 5) and A' is at (5,1)? Wait, no, let's use rotation rules.

Rotation rules:

- 90° clockwise: (x, y) → (y, -x)
- 90° counterclockwise: (x, y) → (-y, x)
- 180°: (x, y) → (-x, -y)
- 270° clockwise (same as 90° counterclockwise): (x, y) → (-y, x)
- 270° counterclockwise (same as 90° clockwise): (x, y) → (y, -x)

Wait, let's find coordinates of A and A'. Let's look at the grid:

A: x = -1, y = 5 (since it's 1 unit left of y-axis, 5 units up)

A': x = 5, y = 1 (5 units right, 1 unit up)

Wait, no, maybe I made a mistake. Wait, maybe A is at (-1, 5) and A' is at (5, 1)? Wait, no, let's check the rotation.

Wait, maybe A is at (-1, 5) and A' is at (5, 1)? Wait, no, let's try 90° clockwise: (x, y) → (y, -x). So A(-1, 5) → (5, 1). Yes! Because y=5, -x = 1 (since x=-1, -x=1). So (5, 1), which matches A'. So 90° clockwise rotation: (-1,5) → (5,1). Correct.

Now 270° counterclockwise: 270° counterclockwise is same as 90° clockwise? Wait, no: 270° counterclockwise rotation: (x, y) → (y, -x), same as 90° clockwise. Wait, 270° counterclockwise: (x, y) → (y, -x). So yes, 270° counterclockwise would also map (-1,5) to (5,1). Wait, no: 270° counterclockwise is (x, y) → (y, -x), same as 90° clockwise. Wait, 90° clockwise: (x,y)→(y, -x). 270° counterclockwise: (x,y)→(y, -x). So both 90° clockwise and 270° counterclockwise give (y, -x).

Wait, let's check 90° clockwise: (-1,5) → (5, 1) (since y=5, -x=1). Correct, because x=-1, so -x=1. So (5,1), which is A'.

Now 270° counterclockwise: same as 90° clockwise, so (x,y)→(y, -x). So (-1,5)→(5,1), same as A'.

Wait, also check 270° clockwise: (x,y)→(-y, x). So (-1,5)→(-5, -1), which is not A'.

90° counterclockwise: (x,y)→(-y, x). So (-1,5)→(-5, -1), not A'.

180°: (x,y)→(-x, -y). So (-1,5)→(1, -5), not A'.

Wait, but maybe I misread the coordinates. Wait, maybe A is at (-1, 5) and A' is at (5,1). So 90° clockwise: (x,y)→(y, -x) → (5, 1), which matches. 270° counterclockwise: (x,y)→(y, -x) (since 270° counterclockwise is same as 90° clockwise). Wait, 270° counterclockwise rotation: the formula is (x, y) → (y, -x), same as 90° clockwise. So 90° clockwise and 270° counterclockwise.

Wait, but let's check again. Wait, maybe A is at (-1, 5) and A' is at (5,1). So 90° clockwise: (x,y)→(y, -x) → (5, 1). Correct. 270° counterclockwise: (x,y)→(y, -x) (since 270° counterclockwise is 3*90° counterclockwise, but the formula for 270° counterclockwise is (x,y)→(y, -x), same as 90° clockwise). Wait, no, let's recall the rotation formulas:

- 90° clockwise: (x, y) → (y, -x)
- 90° counterclockwise: (x, y) → (-y, x)
- 180°: (x, y) → (-x, -y)
- 270° clockwise: (x, y) → (-y, x) (same as 90° counterclockwise)
- 270° counterclockwise: (x, y) → (y, -x) (same as 90° clockwise)

Ah, right! 270° counterclockwise is the same as 90° clockwise. So (x, y) → (y, -x).

So for A(-1, 5):

- 90° clockwise: (5, -(-1)) = (5, 1) → matches A' (5,1)
- 270° counterclockwise: (5, -(-1)) = (5, 1) → matches A'

Now check other rotations:

- 90° counterclockwise: (-5, -1) → not (5,1)
- 180°: (1, -5) → not (5,1)
- 270° clockwise: (-5, -1) → not (5,1)

So the correct rotations are 90° clockwise (A) and 270° counterclockwise (E).

Wait, let's confirm coordinates again. Maybe I misread A's coordinates. Let's look at the graph:

A is at (-1, 5)? Wait, no, the blue dot A: x = -1, y = 5? Wait, the grid: each square is 1 unit. So from origin (0,0), moving left 1 unit (x=-1), up 5 units (y=5). A' is at (5,1): right 5 units, up 1 unit. Yes, so (x,y)=(-1,5) → (y, -x)=(5, 1), which is A'. So 90° clockwise (A) and 270° counterclockwise (E) are correct.

Step2: Verify each option

- A: 90° clockwise → (x,y)→(y, -x) → (-1,5)→(5,1) ✔️
- B: 90° counterclockwise → (-5, -1) ❌
- C: 180° → (1, -5) ❌
- D: 270° clockwise → (-5, -1) ❌
- E: 270° counterclockwise → (y, -x)→(5,1) ✔️

Answer:

A. \( 90^\circ \) clockwise, E. \( 270^\circ \) counterclockwise