Question

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

Explanation:

Step1: Identify original coordinates

First, find the coordinates of point \( K \). From the graph, \( K \) is at \( (-7, 4) \) (assuming the grid lines: x=-7, y=4). Wait, let's check the grid again. Wait, looking at the x-axis: the points N and K are at x=-7? Wait, no, let's see the grid. The x-coordinate for K: between -8 and -6, so x=-7? Wait, no, the grid lines: each square is 1 unit. Let's check the original points:

- \( L \) is at \( (0, 4) \)
- \( M \) is at \( (0, 6) \)
- \( N \) is at \( (-7, 6) \)? Wait, no, wait the horizontal line from N to M: M is at (0,6), N is at (-7,6)? Wait, no, the distance from N to M: let's count the grid squares. From x=-7 to x=0: 7 units? Wait, maybe I misread. Wait, the x-axis: the labels are -10, -8, -6, -4, -2, 0, 2, etc. So each major grid line is 2 units? No, wait, the distance between -10 and -8 is 2 units, so each small square is 1 unit. So from x=-7 (between -8 and -6) to x=0: 7 units? Wait, no, let's look at point K: it's on the horizontal line with L (0,4) and K is at x=-7? Wait, no, maybe the coordinates are:

Wait, let's re-express:

Looking at the graph:

- Point \( L \): (0, 4)
- Point \( M \): (0, 6)
- Point \( N \): (-7, 6)
- Point \( K \): (-7, 4)

Wait, no, maybe the x-coordinate for K is -7? Wait, no, let's check the grid. The vertical line for K: x=-7? Wait, the x-axis labels: -10, -8, -6, -4, -2, 0. So between -8 and -6 is -7. So K is at (-7, 4), N is at (-7, 6), M is at (0, 6), L is at (0, 4).

Now, the rule for 90° clockwise rotation around the origin: the transformation is \( (x, y) ightarrow (y, -x) \).

So for a point \( (x, y) \), after 90° clockwise rotation, the new coordinates \( (x', y') \) are \( (y, -x) \).

Step2: Apply rotation formula

For point \( K(-7, 4) \):

Using the formula \( (x, y) ightarrow (y, -x) \):

\( x' = y = 4 \)

\( y' = -x = -(-7) = 7 \)

Wait, wait, no: the formula for 90° clockwise rotation is \( (x, y) \mapsto (y, -x) \). Let's confirm:

Yes, the rotation matrix for 90° clockwise is \( \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \), so applying to vector \( \begin{pmatrix} x \\ y \end{pmatrix} \), we get \( \begin{pmatrix} y \\ -x \end{pmatrix} \).

So for \( K(-7, 4) \):

\( x' = 4 \)

\( y' = -(-7) = 7 \)

Wait, but let's check again. Wait, maybe the original coordinates are different. Wait, maybe I misread the x-coordinate of K. Let's look again: the horizontal line from K to L: L is at (0,4), K is at (-7,4)? Wait, no, maybe the x-coordinate is -7? Wait, no, maybe the x-coordinate is -7? Wait, let's count the grid squares. From x=0 (L) to x=-7 (K): 7 units to the left. So K is (-7, 4).

Then, applying 90° clockwise rotation: \( (x, y) ightarrow (y, -x) \). So \( (-7, 4) \) becomes \( (4, 7) \)? Wait, no: wait, the formula for 90° clockwise rotation around the origin is \( (x, y) ightarrow (y, -x) \). Wait, let's verify with a simple point: (1,0) rotated 90° clockwise is (0,-1)? No, wait, no: 90° clockwise rotation of (1,0) around origin: the point (1,0) moves to (0,-1)? Wait, no, that's 270° clockwise. Wait, I think I mixed up the formula.

Wait, correct formula for 90° clockwise rotation around origin: \( (x, y) ightarrow (y, -x) \). Wait, no, let's recall:

- 90° counterclockwise: \( (x, y) ightarrow (-y, x) \)
- 90° clockwise: \( (x, y) ightarrow (y, -x) \)

Wait, let's test with (1,0): 90° clockwise: (0, -1)? No, that's not right. Wait, no, (1,0) rotated 90° clockwise around origin: the point (1,0) is on the x-axis. Rotating 90° clockwise would move it to the y-axis, downward. So (0, -1). Using the formula \( (x, y) ightarrow (y, -x) \): (1,0) becomes (0, -1), which is correct.

Another test: (0,1) rotated 90° clockwise: (1, 0). Using formula: (0,1) → (1, 0), correct.

Another test: (1,1) rotated 90° clockwise: (1, -1)? Wait, no: (1,1) rotated 90° clockwise: the image should be (1, -1)? Wait, no, let's plot: (1,1) is in first quadrant. Rotating 90° clockwise: it moves to (1, -1)? No, that's not right. Wait, no, 90° clockwise rotation of (x,y) is (y, -x). So (1,1) → (1, -1). Wait, but when you rotate (1,1) 90° clockwise around origin, the new point should be (1, -1)? Let's see: the vector (1,1) rotated 90° clockwise: the rotation matrix for 90° clockwise is \( \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \). So multiplying \( \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \), which is (1, -1). Correct.

So back to point K. Let's find the correct original coordinates. Let's re-examine the graph:

- Point \( L \): (0, 4) (on y-axis, x=0, y=4)
- Point \( M \): (0, 6) (x=0, y=6)
- Point \( N \): Let's see, the horizontal line from N to M: M is at (0,6), N is at (x,6). The vertical line from N to K: K is at (x,4). So the x-coordinate for N and K is the same. Looking at the x-axis: the distance from N to M is 7 units? Wait, from x=-7 to x=0: 7 units. So N is at (-7, 6), K is at (-7, 4). So original coordinates of K: (-7, 4).

Now, apply 90° clockwise rotation: \( (x, y) ightarrow (y, -x) \). So for K(-7, 4):

\( x' = y = 4 \)

\( y' = -x = -(-7) = 7 \)

Wait, but that would place K' at (4, 7). But let's check with the other points to confirm.

Point L(0,4): rotated 90° clockwise: (4, 0) (since x=0, y=4: (4, -0) = (4,0))

Point M(0,6): rotated 90° clockwise: (6, 0)

Point N(-7,6): rotated 90° clockwise: (6, 7) (x=-7, y=6: (6, -(-7)) = (6,7))

Wait, but let's check the rotation direction. Wait, maybe I made a mistake in the formula. Let's use the correct rotation matrix for 90° clockwise: \( R_{90^\circ \text{ clockwise}} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \). So for a point \( (x, y) \), the new coordinates are \( (y, -x) \).

So for K(-7, 4):

\( x' = 4 \)

\( y' = -(-7) = 7 \)

So K' is (4, 7). Wait, but let's visualize: rotating a point (-7,4) 90° clockwise around origin. The original point is in the second quadrant (x negative, y positive). Rotating 90° clockwise: it should move to the first quadrant (x positive, y positive)? Wait, no: 90° clockwise rotation of a second quadrant point (x negative, y positive) would have x = y (positive, since y is positive) and y = -x (positive, since x is negative, -x is positive). So yes, (4,7) is in the first quadrant, which makes sense.

Wait, but let's check with another approach. The rotation of a point (x, y) 90° clockwise around the origin can be thought of as (x, y) → (y, -x). So for K(-7, 4):

x' = 4, y' = -(-7) = 7. So K' is (4, 7).

But wait, maybe the original coordinates of K are different. Let's re-examine the graph. Let's look at the x-axis: the point K is on the horizontal line with L(0,4). So the y-coordinate of K is 4. The x-coordinate: looking at the grid, the vertical line for K is at x=-7? Wait, between -8 and -6, so x=-7. So K is (-7, 4).

So applying the rotation formula: (x, y) → (y, -x). So (-7, 4) → (4, 7).

Wait, but maybe I made a mistake in the formula. Let's check with (1, 2) rotated 90° clockwise: (2, -1). Let's plot: (1,2) is in first quadrant. Rotating 90° clockwise: it moves to (2, -1), which is in fourth quadrant. Correct, because 90° clockwise rotation of (1,2) would have x=2, y=-1.

So the formula is correct. Therefore, for K(-7, 4), the rotated coordinates are (4, 7).

Wait, but the problem says "Write the coordinates of the vertices after a rotation 90° clockwise around the origin." So we need to find K', L', M', N'? But the question is about K', as per the last line: \( K'(\square, \square) \).

So let's confirm the original coordinates of K. Let's look at the graph again:

- The x-axis: the labels are -10, -8, -6, -4, -2, 0, 2, etc. So each major grid line is 2 units? No, the distance between -10 and -8 is 2 units, so each small square is 1 unit. So from x=-8 to x=-6 is 2 units, so each small square is 1 unit. So the point K is at x=-7 (between -8 and -6) and y=4. So K(-7, 4).

Applying 90° clockwise rotation: (x, y) → (y, -x). So (-7, 4) → (4, 7).

Wait, but let's check with L(0,4): rotated 90° clockwise is (4, 0). M(0,6) → (6, 0). N(-7,6) → (6, 7). So the rotated figure would have vertices at K'(4,7), L'(4,0), M'(6,0), N'(6,7), which is a rectangle, as expected.

So the coordinates of K' are (4, 7).

Wait, but maybe I misread the original coordinates. Let's check again:

Looking at the graph, the horizontal line from K to L: L is at (0,4), K is at (x,4). The vertical line from K to N: N is at (x,6), M is at (0,6). So the length of KL is the distance from x to 0, which is |x - 0|. The length of KN is 6 - 4 = 2. So KL is, say, 7 units (from x=-7 to 0). So K is (-7, 4).

Yes, so K(-7, 4). Rotated 90° clockwise: (4, 7).

Answer:

\( K'(4, 7) \)