Question

write the coordinates of the vertices after a rotation 90° counterclockwise around the origin.
j((□,□))
k((□,□))
l((□,□))

Explanation:

Step1: Find original coordinates

From the graph:

• \( J(-6, 5) \) (wait, no, looking at the grid, J is at x=-6, y=5? Wait, no, the y-axis: the grid lines, J is at (-6, 5)? Wait, no, the y-coordinate for J: the horizontal line for J is at y=5? Wait, no, the grid: each square is 1 unit. Let's check again. J is at (-6, 5)? Wait, K is at (-1, 5)? Wait, no, the x-axis: origin is (0,0). Let's see: J is at x=-6, y=5? Wait, no, the y-coordinate for J: the line is at y=5? Wait, the graph: J is at (-6, 5)? K is at (-1, 5)? L is at (-2, 10)? Wait, no, let's re-express:

Wait, the x-axis: left of origin is negative, right is positive. Y-axis: up is positive, down is negative.

Looking at J: x=-6, y=5 (since it's 6 units left of origin, 5 units up).

K: x=-1, y=5 (1 unit left of origin, 5 units up).

L: x=-2, y=10 (2 units left of origin, 10 units up).

Step2: Apply 90° counterclockwise rotation formula

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

For point J(-6, 5):

Apply the formula: \( x' = -y = -5 \), \( y' = x = -6 \). So \( J'(-5, -6) \)? Wait, no, wait the formula is \((x, y) \to (-y, x)\). Wait, let's confirm: 90° counterclockwise rotation: (x, y) becomes (-y, x). Yes. So for J(-6, 5):

\( x' = -5 \), \( y' = -6 \)? Wait, no: \( x' = -y = -5 \), \( y' = x = -6 \). So J'(-5, -6)? Wait, no, wait original J: let's check again. Wait, maybe I misread the coordinates. Let's look at the graph again.

Wait, J is at (-6, 5)? Wait, K is at (-1, 5)? Wait, no, the x-coordinate for K: from origin (0,0), moving left 1 unit: x=-1, y=5. L is at (-2, 10) (x=-2, y=10).

Wait, let's re-express:

J: x=-6, y=5 (since it's 6 units left, 5 units up).

K: x=-1, y=5 (1 unit left, 5 units up).

L: x=-2, y=10 (2 units left, 10 units up).

Now apply rotation:

J(-6, 5):

\( (x, y) \to (-y, x) \) → \( (-5, -6) \)? Wait, no: \( x' = -y = -5 \), \( y' = x = -6 \). So J'(-5, -6).

K(-1, 5):

\( (x, y) \to (-y, x) \) → \( (-5, -1) \).

L(-2, 10):

\( (x, y) \to (-y, x) \) → \( (-10, -2) \).

Wait, but let's check the formula again. Wait, 90° counterclockwise rotation: the rule is (x, y) → (-y, x). Yes. So for example, (1, 0) rotated 90° counterclockwise is (0, 1)? Wait, no: (1, 0) rotated 90° counterclockwise around origin is (0, 1)? Wait, no, (1, 0) rotated 90° counterclockwise: the point (1,0) is on x-axis, rotating 90° counterclockwise would be (0,1). Wait, using the formula: (x, y)=(1,0) → (-y, x)= (0, 1). Correct. Another example: (0,1) rotated 90° counterclockwise is (-1, 0). Formula: (0,1) → (-1, 0). Correct. So the formula is correct: \((x, y) \to (-y, x)\).

So let's recheck the original coordinates:

Wait, maybe I made a mistake in J's y-coordinate. Let's look at the graph again. The horizontal line for J and K is at y=5? Wait, the y-axis: the line for J and K is at y=5? So J is at (-6, 5), K at (-1, 5), L at (-2, 10). Yes.

So:

• J(-6, 5): rotate 90° CCW: (-5, -6)
• K(-1, 5): rotate 90° CCW: (-5, -1)
• L(-2, 10): rotate 90° CCW: (-10, -2)

Wait, but let's confirm with another approach. Let's take K(-1,5): applying (x,y)→(-y,x) gives (-5, -1). Let's plot mentally: rotating ( -1,5 ) 90° counterclockwise: the x becomes -y ( -5 ), y becomes x ( -1 ). So yes.

Answer:

\( J'(-5, -6) \), \( K'(-5, -1) \), \( L'(-10, -2) \)

Wait, but wait, maybe I messed up the original coordinates. Let's re-express the original points:

Looking at the graph:

• J: x=-6, y=5 (since it's 6 units left of origin, 5 units up)
• K: x=-1, y=5 (1 unit left, 5 units up)
• L: x=-2, y=10 (2 units left, 10 units up)

Yes. So applying the rotation formula:

J(-6,5) → (-5, -6)

K(-1,5) → (-5, -1)

L(-2,10) → (-10, -2)

So the coordinates are:

\( J'(-5, -6) \), \( K'(-5, -1) \), \( L'(-10, -2) \)