Question

Find the coordinates of Z′ after a reflection across y=1 . (1 point) Responses Z′(6,2) upper Z prime left parenthesis 6 comma 2 right parenthesis Z′(−4,2) upper Z prime left parenthesis negative 4 comma 2 right parenthesis Z′(4,0) upper Z prime left parenthesis 4 comma 0 right parenthesis Z′(−4,0)

Answers

Answered by GPT-4o mini
To find the coordinates of \( Z' \) after a reflection across the line \( y = 1 \), we can use the following method.

1. **Identify the original point**: You haven't specified the original point \( Z \). Let's denote the coordinates of point \( Z \) as \( (x, y) \).

2. **Calculate the reflection**: The line \( y = 1 \) is horizontal, so the y-coordinate of the point will change, while the x-coordinate will remain the same. The distance from the original point to the line \( y = 1 \) is \( y - 1 \). The reflected point \( Z' \) will be the same distance below the line as the original point was above it.

To reflect a point \( (x, y) \) across the line \( y = 1 \):

- If \( y > 1 \):
\[
Z' = (x, 1 - (y - 1)) = (x, 2 - y)
\]

- If \( y < 1 \):
\[
Z' = (x, 1 + (1 - y)) = (x, 2 - y)
\]

In both cases, the formula simplifies to:
\[
Z' = (x, 2 - y)
\]

Assuming you provided the coordinates of \( Z \) as one of the responses to find \( Z' \):

Let's check the given responses by assuming each could be a potential \( (x, y) \):

1. \( Z' (6, 2) \):
Assume \( Z (6, 0) \): The reflection would be \( (6, 2) \). Hence this is a valid reflection since the original point \( Z \) below y = 1 reflects to above.

2. \( Z' (-4, 2) \):
Assume \( Z (-4, 0) \): The reflection would be \( (-4, 2) \). This is valid.

3. \( Z' (4, 0) \):
Assume \( Z (4, 2) \): The reflection would be \( (4, 0) \). This reflects point above line down below.

4. \( Z' (-4, 0) \):
Assume \( Z (-4, 2) \): The reflection would be \( (-4, 0) \). This reflects point above line down below.

Based on the reflection rule, the options \( (6, 2), (-4, 2), (4, 0), (-4, 0) \) are candidates based on varying original coordinates.

If Z = ( ) coordinate determines eventually Z’ we confirm each potential. In the standard context of transformations across \( y = 1 \),

One evident example is:
- If we reflect \( (x, 0) \) through \( y=1 \): each turns into \( (x, 2) \) which represents \( Z' \).

Hence, as a systematic conclusion for reflection across \( y=1 \), one reasonable solution remains \( Z'=(6,2) \).

**Final Answer:**

\( Z' (6, 2) \)
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