Question

Trapezoid WXYZ has vertices W(–1,4), X(2, 3), Y(2, 0), and Z(–4, 2). The trapezoid is translated 4 units left and 1 unit down.

Write the coordinates of the image.

Answer 1

The graph of trapezoid EFGH
has coordinates E(–3
, 3)
, F(–1
, 1)
, G(–1
, –1)
and H(–3
, –1)
.

Graph trapezoid EFGH
and its image after a reflection across the y-axis.

Answer 2

Trapezoids E F G H and E prime F prime G prime H prime. E prime (1, 3), F prime (3, 1), G prime (3, negative 1), H prime (1, negative 1).

Trapezoids E F G H and E prime F prime G prime H prime. E prime (1, negative 3), F prime (3, negative 1), G prime (3, 1), H prime (1, 1).

Trapezoids E F G H and E prime F prime G prime H prime. E prime (3, 3), F prime (1, 1), G prime (1, negative 1), H prime (3, negative 1).

Trapezoids E F G H and E prime F prime G prime H prime. E prime (1, negative 1), F prime (3, negative 3), G prime (3, negative 5), H prime (1, negative 5).

Answer 3

The graph of trapezoid EFGH
has coordinates E(–3
, 3)
, F(–1
, 1)
, G(–1
, –1)
and H(–3
, –1)
.

Graph trapezoid EFGH
and its image after a reflection across the y-axis.



Multiple choice question.

Trapezoids E F G H and E prime F prime G prime H prime. E prime (1, 3), F prime (3, 1), G prime (3, negative 1), H prime (1, negative 1).


Trapezoids E F G H and E prime F prime G prime H prime. E prime (1, negative 3), F prime (3, negative 1), G prime (3, 1), H prime (1, 1).


Trapezoids E F G H and E prime F prime G prime H prime. E prime (3, 3), F prime (1, 1), G prime (1, negative 1), H prime (3, negative 1).


Trapezoids E F G H and E prime F prime G prime H prime. E prime (1, negative 1), F prime (3, negative 3), G prime (3, negative 5), H prime (1, negative 5).

Answer 4

To find the coordinates of the image of trapezoid WXYZ after it is translated 4 units left and 1 unit down, we will apply the translation to each vertex.

The translation rules for moving a point (x, y) are:
- For moving 4 units left, we subtract 4 from the x-coordinate: \( x' = x - 4 \)
- For moving 1 unit down, we subtract 1 from the y-coordinate: \( y' = y - 1 \)

Now we will apply this to each vertex of trapezoid WXYZ:

1. **Vertex W(–1, 4):**
- New x-coordinate: \( -1 - 4 = -5 \)
- New y-coordinate: \( 4 - 1 = 3 \)
- New coordinates: W'(-5, 3)

2. **Vertex X(2, 3):**
- New x-coordinate: \( 2 - 4 = -2 \)
- New y-coordinate: \( 3 - 1 = 2 \)
- New coordinates: X'(-2, 2)

3. **Vertex Y(2, 0):**
- New x-coordinate: \( 2 - 4 = -2 \)
- New y-coordinate: \( 0 - 1 = -1 \)
- New coordinates: Y'(-2, -1)

4. **Vertex Z(–4, 2):**
- New x-coordinate: \( -4 - 4 = -8 \)
- New y-coordinate: \( 2 - 1 = 1 \)
- New coordinates: Z'(-8, 1)

Now, we can summarize the new vertices:

- W'(-5, 3)
- X'(-2, 2)
- Y'(-2, -1)
- Z'(-8, 1)

Thus, the coordinates of the image of trapezoid WXYZ after the translation are:
- W'(-5, 3), X'(-2, 2), Y'(-2, -1), Z'(-8, 1)

Answer 5

To graph trapezoid EFGH with the vertices:

- **E**(-3, 3)
- **F**(-1, 1)
- **G**(-1, -1)
- **H**(-3, -1)

### Step 1: Graph the original trapezoid EFGH

1. **Plot the points** on a coordinate plane:
- E(-3, 3): This point is 3 units up from the x-axis and 3 units left from the y-axis.
- F(-1, 1): This point is 1 unit up from the x-axis and 1 unit left from the y-axis.
- G(-1, -1): This point is 1 unit down from the x-axis and 1 unit left from the y-axis.
- H(-3, -1): This point is 1 unit down from the x-axis and 3 units left from the y-axis.

2. **Connect the points** in the order E, F, G, H to form trapezoid EFGH. Note that EF and HG are the parallel sides, which is characteristic of a trapezoid.

### Step 2: Reflect the trapezoid across the y-axis

To find the coordinates of the reflected trapezoid E'F'G'H', we will use the reflection rule across the y-axis, which changes the x-coordinate of each point to its opposite:

- E(-3, 3) goes to E'(3, 3) (y-coordinate remains the same, x-coordinate changes sign)
- F(-1, 1) goes to F'(1, 1)
- G(-1, -1) goes to G'(1, -1)
- H(-3, -1) goes to H'(3, -1)

### Step 3: Graph the reflected trapezoid E'F'G'H'

1. **Plot the new points**:
- E'(3, 3): This point is 3 units up from the x-axis and 3 units right from the y-axis.
- F'(1, 1): This point is 1 unit up from the x-axis and 1 unit right from the y-axis.
- G'(1, -1): This point is 1 unit down from the x-axis and 1 unit right from the y-axis.
- H'(3, -1): This point is 1 unit down from the x-axis and 3 units right from the y-axis.

2. **Connect the points** in the order E', F', G', H' to form the reflected trapezoid E'F'G'H'.

### Summary of Coordinates
- Original trapezoid vertices:
- E(-3, 3)
- F(-1, 1)
- G(-1, -1)
- H(-3, -1)

- Reflected trapezoid vertices:
- E'(3, 3)
- F'(1, 1)
- G'(1, -1)
- H'(3, -1)

By following these steps, you would be able to accurately depict trapezoid EFGH and its reflection across the y-axis.

Answer 6

It appears that you've provided a number of different configurations for the coordinates of the reflected trapezoid \( E'F'G'H' \) based on some transformations from trapezoid \( EFGH \). Let's clarify and confirm the relationship between both trapezoids based on the reflection of \( EFGH \) across the y-axis.

### Original Trapezoid \( EFGH \)
- \( E(-3, 3) \)
- \( F(-1, 1) \)
- \( G(-1, -1) \)
- \( H(-3, -1) \)

### Reflecting \( EFGH \) Across the Y-Axis
To find \( E'F'G'H' \), we reflect each point across the y-axis:
- \( E(-3, 3) \) reflects to \( E'(3, 3) \)
- \( F(-1, 1) \) reflects to \( F'(1, 1) \)
- \( G(-1, -1) \) reflects to \( G'(1, -1) \)
- \( H(-3, -1) \) reflects to \( H'(3, -1) \)

Thus, the reflected coordinates are:
- \( E'(3, 3) \)
- \( F'(1, 1) \)
- \( G'(1, -1) \)
- \( H'(3, -1) \)

### Analyzing the Provided Configurations
Now, let's match the reflected trapezoid coordinates you provided with the computed ones:

1. **\( E' (1, 3), F' (3, 1), G' (3, -1), H' (1, -1) \)**
- This is **not** the correct reflection.

2. **\( E' (1, -3), F' (3, -1), G' (3, 1), H' (1, 1) \)**
- This is also **not** the correct reflection.

3. **\( E' (3, 3), F' (1, 1), G' (1, -1), H' (3, -1) \)**
- This is exactly the reflected trapezoid \( E'F'G'H' \) we calculated.

4. **\( E' (1, -1), F' (3, -3), G' (3, -5), H' (1, -5) \)**
- This is **not** the correct reflection.

### Conclusion
The correct reflection given the vertices of trapezoid \( EFGH \) is the third setup:
- \( E' (3, 3), F' (1, 1), G' (1, -1), H' (3, -1) \).

This matches the reflection of trapezoid \( EFGH \) across the y-axis based on its original coordinates.

Answer 7

To find the correct coordinates of the image of trapezoid \( EFGH \) after reflecting it across the y-axis, we start with the original coordinates:

### Original Trapezoid \( EFGH \)
- \( E(-3, 3) \)
- \( F(-1, 1) \)
- \( G(-1, -1) \)
- \( H(-3, -1) \)

### Reflection Across the Y-Axis
To reflect a point \( (x, y) \) across the y-axis, the x-coordinate becomes its opposite \( (-x, y) \).

1. For **E(-3, 3)**:
- Reflection \( E' = (3, 3) \)

2. For **F(-1, 1)**:
- Reflection \( F' = (1, 1) \)

3. For **G(-1, -1)**:
- Reflection \( G' = (1, -1) \)

4. For **H(-3, -1)**:
- Reflection \( H' = (3, -1) \)

### Reflected Coordinates
So the reflected trapezoid \( E'F'G'H' \) has coordinates:
- \( E'(3, 3) \)
- \( F'(1, 1) \)
- \( G'(1, -1) \)
- \( H'(3, -1) \)

### Comparing with the Multiple Choices
Now, let’s check the provided multiple-choice options:

1. **\( E' (1, 3), F' (3, 1), G' (3, -1), H' (1, -1) \)**
- This is **not** correct.

2. **\( E' (1, -3), F' (3, -1), G' (3, 1), H' (1, 1) \)**
- This is **not** correct.

3. **\( E' (3, 3), F' (1, 1), G' (1, -1), H' (3, -1) \)**
- This is exactly correct!

4. **\( E' (1, -1), F' (3, -3), G' (3, -5), H' (1, -5) \)**
- This is **not** correct.

### Conclusion
The correct answer, which matches our calculated reflected coordinates for trapezoid \( EFGH \), is:

**Trapezoids \( EFGH \) and \( E'F'G'H' \): \( E' (3, 3), F' (1, 1), G' (1, -1), H' (3, -1) \)**