Question

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

b((□,□))
c((□,□))
d((□,□))
e((□,□))

Explanation:

Step1: Find original coordinates

First, identify the original coordinates of points B, C, D, E from the graph.
- Point B: Looking at the grid, B is at \((-7, 3)\) (x=-7, y=3)
- Point C: C is at \((-6, 3)\) (x=-6, y=3)
- Point D: D is at \((-6, 4)\) (x=-6, y=4)
- Point E: E is at \((-7, 4)\) (x=-7, y=4)

Step2: Apply 90° counterclockwise rotation rule

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

For point B \((-7, 3)\):

Apply the rule: \(x=-7\), \(y=3\) → new \(x = -y = -3\), new \(y = x = -7\) → \(B'(-3, -7)\)? Wait, no, wait: Wait, original coordinates: Wait, let's recheck the grid. Wait, the x-axis: from the graph, the square is at x around -7 and -6? Wait, looking at the grid, the x-coordinate for B: let's count the grid lines. The origin is (0,0). Moving left from origin, each grid is 1 unit. So B: x=-7? Wait, no, the x-axis labels: -10, -8, -6, -4, -2, 0, 2... So between -8 and -6, the x-coordinate for B: let's see, B is at x=-7? Wait, no, the square: B, C, D, E. Let's see the coordinates:

Wait, looking at the graph:

- B: x=-7? Wait, no, the x-axis: the vertical lines are x=-10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2,... So B is at x=-7, y=3? Wait, no, the y-axis: y=2, 3, 4, 5... So B is at (x=-7, y=3)? Wait, no, let's check again. Wait, the square: B is below E, left of C. So:

Wait, E is at (x=-7, y=4), B is at (x=-7, y=3), C is at (x=-6, y=3), D is at (x=-6, y=4). Yes, that's correct. So original coordinates:

- B: \((-7, 3)\)
- C: \((-6, 3)\)
- D: \((-6, 4)\)
- E: \((-7, 4)\)

Now apply the 90° counterclockwise rotation rule \((x, y) \to (-y, x)\):

For B \((-7, 3)\):

\(x = -7\), \(y = 3\) → new \(x = -y = -3\), new \(y = x = -7\) → \(B'(-3, -7)\)? Wait, no, that can't be. Wait, maybe I mixed up the rule. Wait, no: the correct rule for 90° counterclockwise is \((x, y) \mapsto (-y, x)\). Wait, let's take a simple point, like (1, 0). Rotating 90° counterclockwise around origin should be (0, 1). Let's apply the rule: (1,0) → (-0, 1) = (0,1). Correct. Another example: (0,1) → (-1, 0). Correct. (1,1) → (-1, 1). Wait, no: (1,1) rotated 90° counterclockwise is (-1,1)? Wait, no, (1,1) rotated 90° counterclockwise: the rotation matrix is \(\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}\), so multiplying by \(\begin{bmatrix}x \\ y\end{bmatrix}\) gives \(\begin{bmatrix}-y \\ x\end{bmatrix}\). So (1,1) → (-1,1). Correct. So the rule is correct.

Wait, but in our case, the original points are in the second quadrant (x negative, y positive). So rotating 90° counterclockwise:

For B \((-7, 3)\):

\(x = -7\), \(y = 3\) → new \(x = -y = -3\), new \(y = x = -7\) → \(B'(-3, -7)\)? Wait, but that would be in the third quadrant. Wait, but let's check the grid again. Wait, maybe I made a mistake in the original coordinates. Let's re-express the original coordinates correctly.

Wait, looking at the graph:

- The x-coordinate for B: the vertical line (x-axis) at x=-7? Wait, the x-axis labels: -10, -8, -6, -4, -2, 0... Wait, between -8 and -6, there's -7. So B is at x=-7, y=3 (since y=3 is the horizontal line). C is at x=-6, y=3. D is at x=-6, y=4. E is at x=-7, y=4. Yes.

So applying the rule:

- B \((-7, 3)\) → \((-y, x)\) → \(x=-3\), \(y=-7\) → \(B'(-3, -7)\)
- C \((-6, 3)\) → \((-y, x)\) → \(x=-3\), \(y=-6\) → \(C'(-3, -6)\)? Wait, no: Wait, \(x=-6\), \(y=3\) → new \(x = -y = -3\), new \(y = x = -6\) → \(C'(-3, -6)\)
- D \((-6, 4)\) → \((-y, x)\) → \(x=-4\), \(y=-6\) → \(D'(-4, -6)\)
- E \((-7, 4)\) → \((-y, x)\) → \(x=-4\), \(y=-7\) → \(E'(-4, -7)\)

Wait, but that seems odd. Wait, maybe I mixed up the rotation direction. Wait, 90° counterclockwise: the formula is (x,y) → (-y, x). Let's confirm with a point in the second quadrant: ( -2, 1 ). Rotating 90° counterclockwise: should be (-1, -2). Let's apply the rule: (x=-2, y=1) → (-y=-1, x=-2) → (-1, -2). Correct. So yes, the rule is correct.

Wait, but let's check the original coordinates again. Wait, maybe the original coordinates are different. Wait, looking at the graph, the square is at x around -7 and -6, y around 3 and 4. So:

B: (-7, 3)

C: (-6, 3)

D: (-6, 4)

E: (-7, 4)

Applying the 90° counterclockwise rotation:

B: (-7,3) → (-3, -7)

C: (-6,3) → (-3, -6)

D: (-6,4) → (-4, -6)

E: (-7,4) → (-4, -7)

Wait, but let's check with another approach. Let's plot mentally. The original square is in the second quadrant (x negative, y positive). Rotating 90° counterclockwise around origin: the square should move to the third quadrant? Wait, no: 90° counterclockwise from second quadrant (x-, y+) would be third quadrant (x-, y-)? Wait, no: the rotation direction: 90° counterclockwise from (x,y) in second quadrant (x negative, y positive) → the new x is -y (which is negative, since y is positive), new y is x (which is negative, since x is negative). So yes, third quadrant.

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

Wait, the x-axis: each grid line is 1 unit. So from the origin (0,0) to the left, x=-1, -2, ..., -7, -8, etc. The y-axis: from origin up, y=1,2,3,4, etc. So the square:

- B: x=-7, y=3 (so (x=-7, y=3))

- C: x=-6, y=3 (so (x=-6, y=3))

- D: x=-6, y=4 (so (x=-6, y=4))

- E: x=-7, y=4 (so (x=-7, y=4))

Applying the rotation rule \((x, y) \to (-y, x)\):

- B: (-7, 3) → (-3, -7) → B'(-3, -7)

- C: (-6, 3) → (-3, -6) → C'(-3, -6)

- D: (-6, 4) → (-4, -6) → D'(-4, -6)

- E: (-7, 4) → (-4, -7) → E'(-4, -7)

Wait, but let's check with a different rule. Wait, maybe the rule is (x,y) → (y, -x) for 90° clockwise, and (-y, x) for 90° counterclockwise. Yes, that's correct. So 90° counterclockwise: (-y, x).

So the coordinates after rotation are:

B'(-3, -7)

C'(-3, -6)

D'(-4, -6)

E'(-4, -7)

Wait, but let's verify with a simple point. Take (0,1), rotate 90° counterclockwise: ( -1, 0 ). Apply the rule: (x=0, y=1) → (-y=-1, x=0) → (-1, 0). Correct. Take (1,0), rotate 90° counterclockwise: (0,1). Apply rule: (x=1, y=0) → (-0=0, x=1) → (0,1). Correct. Take (-1, 0), rotate 90° counterclockwise: (0, -1). Apply rule: (x=-1, y=0) → (-0=0, x=-1) → (0, -1). Correct. Take (0, -1), rotate 90° counterclockwise: (1, 0). Apply rule: (x=0, y=-1) → (-(-1)=1, x=0) → (1, 0). Correct. So the rule is correct.

So applying the rule to each point:

- B: (-7, 3) → (-3, -7)

- C: (-6, 3) → (-3, -6)

- D: (-6, 4) → (-4, -6)

- E: (-7, 4) → (-4, -7)

Answer:

\(B'(-3, -7)\), \(C'(-3, -6)\), \(D'(-4, -6)\), \(E'(-4, -7)\)

Wait, but let's check again. Wait, maybe the original coordinates are (x=-7, y=3) is wrong. Wait, looking at the graph, the square is at x=-7 and x=-6? Wait, the x-axis labels: -10, -8, -6, -4, -2, 0... So between -8 and -6, there's -7. So B is at x=-7, y=3. Yes. So the rotation is correct.