Question

write the coordinates of the vertices after a rotation 90° counterclockwise around the origin.
j(\boxed{\quad}, \boxed{\quad})
k(\boxed{\quad}, \boxed{\quad})
l(\boxed{\quad}, \boxed{\quad})
m(\boxed{\quad}, \boxed{\quad})

Explanation:

Step1: Find original coordinates

First, identify the original coordinates of each vertex from the graph.

• \( J \): Looking at the grid, \( J \) is at \( (-9, 8) \) (x=-9, y=8)
• \( K \): \( K \) is at \( (-6, 8) \) (x=-6, y=8)
• \( L \): \( L \) is at \( (-6, 9) \) (x=-6, y=9)
• \( M \): \( M \) is at \( (-9, 9) \) (x=-9, y=9)

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

Substitute \( x = -9 \), \( y = 8 \) into the rule: \( (-y, x) = (-8, -9) \). So \( J'(-8, -9) \).

For \( K(-6, 8) \):

Substitute \( x = -6 \), \( y = 8 \) into the rule: \( (-y, x) = (-8, -6) \). So \( K'(-8, -6) \).

For \( L(-6, 9) \):

Substitute \( x = -6 \), \( y = 9 \) into the rule: \( (-y, x) = (-9, -6) \). So \( L'(-9, -6) \).

For \( M(-9, 9) \):

Substitute \( x = -9 \), \( y = 9 \) into the rule: \( (-y, x) = (-9, -9) \). Wait, no: Wait, \( x=-9 \), \( y=9 \), so \( -y = -9 \), \( x = -9 \)? Wait, no, wait: Wait, the rule is \( (x,y) \to (-y, x) \). So for \( M(-9,9) \): \( -y = -9 \), \( x = -9 \)? Wait, no, \( x=-9 \), \( y=9 \), so \( -y = -9 \), \( x = -9 \)? Wait, that would be \( (-9, -9) \)? Wait, no, wait, let's recheck. Wait, original \( M \): x=-9, y=9. So applying \( (x,y) \to (-y, x) \): \( -y = -9 \), \( x = -9 \)? Wait, no, \( x \) is -9, so the new x is -y (which is -9), new y is x (which is -9)? Wait, that seems off. Wait, maybe I made a mistake in original coordinates. Wait, looking at the graph again: Let's re-express the coordinates. Let's count the grid lines. The x-axis: from -10 to 10, y-axis from -10 to 10. Let's check the x-coordinates: For J, the x is -9? Wait, maybe I miscounted. Wait, the vertical lines: each grid is 1 unit. Let's see, the point J: looking at the x-axis, the first vertical line left of -8 is -9? Wait, no, maybe the coordinates are: Let's see, J is at (x=-9, y=8)? Wait, maybe the x-coordinates are: Let's check the horizontal positions. Let's see, the origin is (0,0). Moving left: -1, -2, ..., -9, -10. So J is at x=-9, y=8. K is at x=-6, y=8. L is at x=-6, y=9. M is at x=-9, y=9. So that's correct. Then applying the rotation rule \( (x,y) \to (-y, x) \):

For J(-9,8): new x = -8, new y = -9 → (-8, -9)

For K(-6,8): new x = -8, new y = -6 → (-8, -6)

For L(-6,9): new x = -9, new y = -6 → (-9, -6)

For M(-9,9): new x = -9, new y = -9 → (-9, -9)? Wait, no, wait: \( x=-9 \), \( y=9 \), so \( -y = -9 \), \( x = -9 \). So yes, (-9, -9). Wait, but let's verify with a simpler point. For example, a point (1,2) rotated 90° counterclockwise should be (-2,1). Let's check: (x=1, y=2) → (-y, x) = (-2,1). Correct. So for ( -9,8 ): (-8, -9). Correct. For (-6,8): (-8, -6). Correct. For (-6,9): (-9, -6). Correct. For (-9,9): (-9, -9). Wait, but let's check the graph again. Wait, maybe the original coordinates are different. Wait, maybe J is at (-9, 8), K at (-6,8), L at (-6,9), M at (-9,9). So after rotation:

J'(-8, -9)

K'(-8, -6)

L'(-9, -6)

M'(-9, -9)

Wait, but let's confirm the rotation rule again. The 90° counterclockwise rotation around the origin transforms (x, y) to (-y, x). Yes, that's the standard rule. So that's correct.

Answer:

• \( J'(-8, -9) \)
• \( K'(-8, -6) \)
• \( L'(-9, -6) \)
• \( M'(-9, -9) \)