Question

The image of trapezoid ABCD has coordinates A′(–2, –5), B′(–1, –2), C′(2, –2), and D′(3, –5). It was translated by the rule T–1, –3(x, y). Which diagram shows the pre-image?

On a coordinate plane, a trapezoid has points A (negative 1, negative 2), B (0, 1), C (3, 1), D (4, negative 2).

On a coordinate plane, a trapezoid has points A (negative 3, negative 2), B (negative 2, 1), C (1, 1), D (2, negative 2).

On a coordinate plane, a trapezoid has points A (negative 4, negative 5), B (negative 3, negative 2), C (0, negative 2), D (1, negative 5).

On a coordinate plane, a trapezoid has points A (0, negative 2), B (1, 1), C (4, 1), D (5, negative 2).

Answer 1

To find the pre-image coordinates of trapezoid ABCD after applying the translation rule \( T_{-1, -3}(x, y) \), we need to reverse the translation. The translation rule \( T_{-1, -3}(x, y) \) means that every point (x, y) in the pre-image is transformed to (x - 1, y - 3) in the image.

Therefore, to find the pre-image coordinates A, B, C, and D from the image coordinates A′, B′, C′, and D′, we can apply the following transformations:

- For point A′(–2, –5):
\[
A(-1, -2) = A'(-2 + 1, -5 + 3) = (-1, -2)
\]

- For point B′(–1, –2):
\[
B(0, 1) = B'(-1 + 1, -2 + 3) = (0, 1)
\]

- For point C′(2, –2):
\[
C(3, 1) = C'(2 + 1, -2 + 3) = (3, 1)
\]

- For point D′(3, –5):
\[
D(4, -2) = D'(3 + 1, -5 + 3) = (4, -2)
\]

Thus, the pre-image coordinates are:
- A(–1, –2)
- B(0, 1)
- C(3, 1)
- D(4, –2)

Now, let’s check which of the given trapezoids corresponds to these points:

1. **First option**: A(–1, –2), B(0, 1), C(3, 1), D(4, –2) — This matches our pre-image.

2. **Second option**: A(–3, –2), B(–2, 1), C(1, 1), D(2, –2) — This does not match.

3. **Third option**: A(–4, –5), B(–3, –2), C(0, –2), D(1, –5) — This does not match.

4. **Fourth option**: A(0, –2), B(1, 1), C(4, 1), D(5, –2) — This does not match.

Therefore, the correct diagram that shows the pre-image of trapezoid ABCD is the first option:
- A(–1, –2), B(0, 1), C(3, 1), D(4, –2).