Question

reflection across x = -1
rotation 90° counterclockwise about the origin

Explanation:

To solve problems related to geometric transformations (reflection and rotation) of the given triangles, we analyze the coordinates of the vertices and apply the transformation rules.

Part 1: Reflection across \( x = -1 \)

Let's first determine the coordinates of \( \triangle ABC \):

• \( A(-4, 3) \), \( B(-1, 4) \), \( C(-2, 0) \) (assuming grid units).

The rule for reflecting a point \( (x, y) \) across the vertical line \( x = a \) is:
\[ (x, y)
ightarrow (2a - x, y) \]
For \( x = -1 \), the rule becomes:
\[ (x, y)
ightarrow (2(-1) - x, y) = (-2 - x, y) \]

Step 1: Reflect \( A(-4, 3) \)

\[ x' = -2 - (-4) = 2, \quad y' = 3 \]
Reflected \( A' = (2, 3) \)

Step 2: Reflect \( B(-1, 4) \)

\[ x' = -2 - (-1) = -1, \quad y' = 4 \]
Reflected \( B' = (-1, 4) \) (since \( B \) lies on \( x = -1 \), it maps to itself).

Step 3: Reflect \( C(-2, 0) \)

\[ x' = -2 - (-2) = 0, \quad y' = 0 \]
Reflected \( C' = (0, 0) \)

Part 2: Rotation \( 90^\circ \) counterclockwise about the origin

The rule for rotating a point \( (x, y) \) \( 90^\circ \) counterclockwise about the origin is:
\[ (x, y)
ightarrow (-y, x) \]

Assume the coordinates of \( \triangle DEF \) (before rotation) are, e.g., \( D(2, -2) \), \( E(3, 0) \), \( F(3, -3) \) (from the grid).

Step 1: Rotate \( D(2, -2) \)

\[ x' = -(-2) = 2, \quad y' = 2 \]
Rotated \( D' = (2, 2) \) (Wait, correction: \( 90^\circ \) counterclockwise: \( (x, y)
ightarrow (-y, x) \). So \( D(2, -2) \):
\[ x' = -(-2) = 2, \quad y' = 2 \]? No, wait: \( x = 2 \), \( y = -2 \). So \( -y = 2 \), \( x = 2 \). So \( D' = (2, 2) \).

Step 2: Rotate \( E(3, 0) \)

\[ x' = -0 = 0, \quad y' = 3 \]
Rotated \( E' = (0, 3) \)

Step 3: Rotate \( F(3, -3) \)

\[ x' = -(-3) = 3, \quad y' = 3 \]
Rotated \( F' = (3, 3) \)

Final Answer (Example)

For reflection across \( x = -1 \), the image of \( \triangle ABC \) has vertices \( A'(2, 3) \), \( B'(-1, 4) \), \( C'(0, 0) \).

For \( 90^\circ \) counterclockwise rotation about the origin, the image of \( \triangle DEF \) (with \( D(2, -2) \), \( E(3, 0) \), \( F(3, -3) \)) has vertices \( D'(2, 2) \), \( E'(0, 3) \), \( F'(3, 3) \).

(Note: Coordinates depend on the exact grid placement. Adjust calculations based on the actual coordinates of the triangle vertices.)

Answer:

To solve problems related to geometric transformations (reflection and rotation) of the given triangles, we analyze the coordinates of the vertices and apply the transformation rules.

Part 1: Reflection across \( x = -1 \)

Let's first determine the coordinates of \( \triangle ABC \):

• \( A(-4, 3) \), \( B(-1, 4) \), \( C(-2, 0) \) (assuming grid units).

The rule for reflecting a point \( (x, y) \) across the vertical line \( x = a \) is:
\[ (x, y)
ightarrow (2a - x, y) \]
For \( x = -1 \), the rule becomes:
\[ (x, y)
ightarrow (2(-1) - x, y) = (-2 - x, y) \]

Step 1: Reflect \( A(-4, 3) \)

\[ x' = -2 - (-4) = 2, \quad y' = 3 \]
Reflected \( A' = (2, 3) \)

Step 2: Reflect \( B(-1, 4) \)

\[ x' = -2 - (-1) = -1, \quad y' = 4 \]
Reflected \( B' = (-1, 4) \) (since \( B \) lies on \( x = -1 \), it maps to itself).

Step 3: Reflect \( C(-2, 0) \)

\[ x' = -2 - (-2) = 0, \quad y' = 0 \]
Reflected \( C' = (0, 0) \)

Part 2: Rotation \( 90^\circ \) counterclockwise about the origin

The rule for rotating a point \( (x, y) \) \( 90^\circ \) counterclockwise about the origin is:
\[ (x, y)
ightarrow (-y, x) \]

Assume the coordinates of \( \triangle DEF \) (before rotation) are, e.g., \( D(2, -2) \), \( E(3, 0) \), \( F(3, -3) \) (from the grid).

Step 1: Rotate \( D(2, -2) \)

\[ x' = -(-2) = 2, \quad y' = 2 \]
Rotated \( D' = (2, 2) \) (Wait, correction: \( 90^\circ \) counterclockwise: \( (x, y)
ightarrow (-y, x) \). So \( D(2, -2) \):
\[ x' = -(-2) = 2, \quad y' = 2 \]? No, wait: \( x = 2 \), \( y = -2 \). So \( -y = 2 \), \( x = 2 \). So \( D' = (2, 2) \).

Step 2: Rotate \( E(3, 0) \)

\[ x' = -0 = 0, \quad y' = 3 \]
Rotated \( E' = (0, 3) \)

Step 3: Rotate \( F(3, -3) \)

\[ x' = -(-3) = 3, \quad y' = 3 \]
Rotated \( F' = (3, 3) \)

Final Answer (Example)

For reflection across \( x = -1 \), the image of \( \triangle ABC \) has vertices \( A'(2, 3) \), \( B'(-1, 4) \), \( C'(0, 0) \).

For \( 90^\circ \) counterclockwise rotation about the origin, the image of \( \triangle DEF \) (with \( D(2, -2) \), \( E(3, 0) \), \( F(3, -3) \)) has vertices \( D'(2, 2) \), \( E'(0, 3) \), \( F'(3, 3) \).

(Note: Coordinates depend on the exact grid placement. Adjust calculations based on the actual coordinates of the triangle vertices.)