Question

graph the image of δabc after a rotation 270° clockwise around the origin.

Explanation:

First, we need to identify the coordinates of points \( A \), \( B \), and \( C \) from the graph. From the grid, we can see:

• Point \( A \): Let's assume the coordinates. Looking at the graph, \( A \) seems to be at \( (3, 5) \) (wait, no, let's check again. Wait, the y-axis: from the grid, \( A \) is at \( (3, 5) \)? Wait, no, the x-axis: the origin is (0,0). Let's re-examine. The x-coordinate: moving right from origin, each grid is 1 unit. The y-coordinate: moving up from origin, each grid is 1 unit. So:
• Point \( C \): Let's see, \( C \) is at \( (1, 4) \) (x=1, y=4)
• Point \( A \): \( (3, 5) \)? Wait, no, the yellow dot for \( A \): x=3, y=5? Wait, the blue dot for \( B \): x=3, y=6? Wait, maybe I misread. Wait, the grid: let's count the units. From origin (0,0), moving right: x=1,2,3,4,... Moving up: y=1,2,3,4,5,6,7,8,... So:
• Point \( C \): x=1, y=4 (since it's 1 unit right on x, 4 units up on y)
• Point \( A \): x=3, y=5 (3 units right, 5 units up)
• Point \( B \): x=3, y=6 (3 units right, 6 units up)

Now, the rule for rotating a point \( (x, y) \) 270° clockwise around the origin is: \( (x, y)
ightarrow (y, -x) \). Wait, no: wait, the rotation rules:

• 90° clockwise: \( (x, y)

ightarrow (y, -x) \)

• 180° clockwise: \( (x, y)

ightarrow (-x, -y) \)

• 270° clockwise: \( (x, y)

ightarrow (-y, x) \)

Wait, let's confirm the rotation rule for 270° clockwise (or equivalently 90° counterclockwise). The correct rule is: when rotating a point \( (x, y) \) 270° clockwise about the origin, the new coordinates \( (x', y') \) are given by \( x' = y \), \( y' = -x \)? Wait, no, let's use the standard rotation matrix. The rotation matrix for \( \theta \) degrees clockwise is \(

$$\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$$

\). For \( \theta = 270^\circ \), \( \cos 270^\circ = 0 \), \( \sin 270^\circ = -1 \). So the matrix is \(

$$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$

\). So applying this to a point \( (x, y) \), we get:

\( x' = 0 \cdot x + (-1) \cdot y = -y \)

\( y' = 1 \cdot x + 0 \cdot y = x \)

So the rule is \( (x, y)
ightarrow (-y, x) \) for 270° clockwise rotation.

Now let's find the coordinates of each point:

1. Point \( C \): Let's confirm its original coordinates. From the graph, \( C \) is at (1, 4) (x=1, y=4). Applying the rotation:

\( x' = -y = -4 \)

\( y' = x = 1 \)

So \( C' \) is at (-4, 1)

1. Point \( A \): Let's check its original coordinates. From the graph, \( A \) is at (3, 5) (x=3, y=5). Applying the rotation:

\( x' = -y = -5 \)

\( y' = x = 3 \)

So \( A' \) is at (-5, 3)

1. Point \( B \): Original coordinates (3, 6) (x=3, y=6). Applying the rotation:

\( x' = -y = -6 \)

\( y' = x = 3 \)

So \( B' \) is at (-6, 3)

Now, we need to plot these new points \( A'(-5, 3) \), \( B'(-6, 3) \), and \( C'(-4, 1) \) on the grid and connect them to form the image of \( \triangle ABC \) after the rotation.

Step 1: Identify Original Coordinates

• \( C(1, 4) \)
• \( A(3, 5) \)
• \( B(3, 6) \)

Step 2: Apply 270° Clockwise Rotation Rule (\( (x, y)

ightarrow (-y, x) \))

• For \( C(1, 4) \): \( x' = -4 \), \( y' = 1 \) → \( C'(-4, 1) \)
• For \( A(3, 5) \): \( x' = -5 \), \( y' = 3 \) → \( A'(-5, 3) \)
• For \( B(3, 6) \): \( x' = -6 \), \( y' = 3 \) → \( B'(-6, 3) \)

Step 3: Plot the New Points

Plot \( A'(-5, 3) \), \( B'(-6, 3) \), and \( C'(-4, 1) \) on the coordinate grid and draw the triangle connecting these points.

Answer:

The image of \( \triangle ABC \) after a 270° clockwise rotation around the origin has vertices at \( A'(-5, 3) \), \( B'(-6, 3) \), and \( C'(-4, 1) \). (To graph, plot these points and connect them.)