Question

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

Explanation:

Step1: Find original coordinates

From the graph: \( J(-8, 3) \), \( K(-8, 9) \), \( L(-2, 0) \).

Step2: Apply 90° counterclockwise rotation rule

The rule for a \( 90^\circ \) counterclockwise rotation about the origin is \( (x, y) \to (-y, x) \).

- For \( J(-8, 3) \):
New \( x = -3 \), new \( y = -8 \)? Wait, no: Wait, rule is \( (x,y) \to (-y, x) \). So \( J(-8, 3) \) becomes \( (-3, -8) \)? Wait, no: Wait, \( x=-8 \), \( y=3 \). So \( -y = -3 \), \( x = -8 \)? Wait, no, I messed up. Wait, the correct rule is \( (x, y) \) rotated 90° counterclockwise is \( (-y, x) \). So:

- \( J(-8, 3) \): \( x=-8 \), \( y=3 \). So new \( x = -y = -3 \), new \( y = x = -8 \)? Wait, no, that can't be. Wait, no, let's recheck. The rotation rule: 90° counterclockwise about origin: \( (x, y) \mapsto (-y, x) \). So for \( J(-8, 3) \): \( -y = -3 \), \( x = -8 \)? Wait, no, \( x \) is -8, \( y \) is 3. So \( (-y, x) = (-3, -8) \)? Wait, but let's check the graph. Wait, maybe I got the coordinates wrong. Let's re-express the coordinates:

Looking at the graph:

- \( J \) is at \( x=-8 \), \( y=3 \) (since it's 8 units left on x, 3 up on y).

- \( K \) is at \( x=-8 \), \( y=9 \) (8 left, 9 up).

- \( L \) is at \( x=-2 \), \( y=0 \) (2 left, on x-axis).

Now, applying \( (x, y) \to (-y, x) \):

- For \( J(-8, 3) \): \( -y = -3 \), \( x = -8 \)? Wait, no, \( x \) is -8, \( y \) is 3. So \( (-y, x) = (-3, -8) \)? Wait, that seems off. Wait, maybe the rule is \( (x, y) \to (-y, x) \). Let's take a point (1,0): rotating 90° counterclockwise should be (0,1). Wait, no: (1,0) rotated 90° counterclockwise is (0,1)? Wait, no, (1,0) around origin 90° counterclockwise: the new x is -0 = 0, new y is 1? Wait, no, (x,y) → (-y, x). So (1,0) → (0,1)? Wait, no, (1,0) rotated 90° counterclockwise is (0,1)? Wait, no, actually, (x,y) rotated 90° counterclockwise is (-y, x). So (1,0) → (0,1)? Wait, no, (1,0) rotated 90° counterclockwise: the x becomes -y (which is 0), y becomes x (which is 1). So (0,1). Correct. (0,1) rotated 90° counterclockwise: (-1, 0). Correct. So the rule is correct.

So for \( J(-8, 3) \): \( x=-8 \), \( y=3 \). So new \( x = -y = -3 \), new \( y = x = -8 \). So \( J'(-3, -8) \)? Wait, but that's in the fourth quadrant? Wait, maybe I made a mistake in the original coordinates. Wait, let's check the grid again. The x-axis: left is negative, right is positive. y-axis: up is positive, down is negative.

\( J \): x=-8 (8 units left), y=3 (3 units up). So ( -8, 3 ).

\( K \): x=-8, y=9 (9 units up). So ( -8, 9 ).

\( L \): x=-2, y=0 (on x-axis). So ( -2, 0 ).

Now, applying \( (x, y) \to (-y, x) \):

- \( J(-8, 3) \): \( -y = -3 \), \( x = -8 \). So \( J'(-3, -8) \)? Wait, no, wait: \( x \) is -8, \( y \) is 3. So \( (-y, x) = (-3, -8) \).

- \( K(-8, 9) \): \( -y = -9 \), \( x = -8 \). So \( K'(-9, -8) \)? Wait, no, that can't be. Wait, maybe I mixed up the rule. Wait, the correct rule for 90° counterclockwise rotation is \( (x, y) \mapsto (-y, x) \). Wait, let's take (2,3): rotating 90° counterclockwise should be (-3, 2). Let's verify with a graph: (2,3) is in the first quadrant. Rotating 90° counterclockwise around origin: it should go to (-3, 2), which is in the second quadrant. Yes, that makes sense. So (2,3) → (-3, 2). So the rule is correct.

So for \( J(-8, 3) \): \( x=-8 \), \( y=3 \). So \( -y = -3 \), \( x = -8 \). So \( J'(-3, -8) \)? Wait, but (-3, -8) is in the fourth quadrant? Wait, maybe the original coordinates are different. Wait, maybe I misread the x-coordinate of J. Let's look at the grid: the vertical lines are x=-10, -9, -8, -7, ..., 0, 1, ..., 10. So J is at x=-8 (the 8th line from the origin to the left), y=3 (3rd line up from x-axis). So ( -8, 3 ).

K is at x=-8, y=9 (9th line up). So ( -8, 9 ).

L is at x=-2 (2nd line left), y=0. So ( -2, 0 ).

Now, applying the rotation:

- \( J(-8, 3) \): \( (-y, x) = (-3, -8) \). So \( J'(-3, -8) \).

- \( K(-8, 9) \): \( (-y, x) = (-9, -8) \). So \( K'(-9, -8) \).

- \( L(-2, 0) \): \( (-y, x) = (0, -2) \). So \( L'(0, -2) \).

Wait, but let's check with another approach. Maybe the rule is \( (x, y) \to (y, -x) \) for clockwise, but counterclockwise is \( (-y, x) \). Yes, that's correct.

Alternatively, maybe I made a mistake in the sign. Let's confirm with a point (0,1): rotating 90° counterclockwise should be (-1, 0). Using the rule: \( (-y, x) = (-1, 0) \). Correct. (1,1) rotated 90° counterclockwise: (-1, 1). Wait, no: (1,1) → (-1, 1)? Wait, no, (1,1) rotated 90° counterclockwise: the x becomes -1 ( -y ), y becomes 1 ( x ). So (-1, 1). Which is correct, as (1,1) is in the first quadrant, rotating 90° counterclockwise would be in the second quadrant, x negative, y positive. Correct.

So back to the points:

- \( J(-8, 3) \): \( (-3, -8) \)

- \( K(-8, 9) \): \( (-9, -8) \)

- \( L(-2, 0) \): \( (0, -2) \)

Wait, but let's check the graph again. Maybe the original coordinates are different. Wait, maybe J is at (-8, 3), K at (-8, 9), L at (-2, 0). Then after rotation:

J'(-3, -8), K'(-9, -8), L'(0, -2).

But let's verify with a simpler point. Let's take L(-2, 0). Rotating 90° counterclockwise: (0, -2). Yes, because (x,y)=(-2,0) → (-0, -2) = (0, -2). Correct.

J(-8, 3): (-3, -8). Correct.

K(-8, 9): (-9, -8). Correct.

So the coordinates after rotation are:

\( J'(-3, -8) \), \( K'(-9, -8) \), \( L'(0, -2) \).

Answer:

\( J'(-3, -8) \), \( K'(-9, -8) \), \( L'(0, -2) \)