Question

Find the coordinates of Z′
after a reflection across y=1
.


(1 point)
Responses

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′(6,2)
upper Z prime left parenthesis 6 comma 2 right parenthesis

Z′(−4,0)

Answer 1

To find the coordinates of \( Z' \) after a reflection across the line \( y = 1 \), we start with the original coordinates of point \( Z = (x, y) \). The formula for reflecting a point \( (x, y) \) across the line \( y = k \) is:

\[
(x, y') = (x, 2k - y)
\]

For our case, \( k = 1 \):

1. The reflection gives us the new y-coordinate:
\[
y' = 2 \cdot 1 - y = 2 - y
\]

Now, we can determine the reflections for the provided coordinates if we assume a generic point. Let's evaluate the four points you've provided to determine where \( Z \) could reasonably be.

1. **For \( Z'(-4, 2) \)**:
\[
y' = 2 - 2 = 0 \quad \rightarrow \text{ This does not match. }
\]

2. **For \( Z'(4, 0) \)**:
\[
y' = 2 - 0 = 2 \quad \rightarrow \text{ This gives back (4, 2). }
\]

3. **For \( Z'(6, 2) \)**:
\[
y' = 2 - 2 = 0 \quad \rightarrow \text{ This does not match. }
\]

4. **For \( Z'(-4, 0) \)**:
\[
y' = 2 - 0 = 2 \quad \rightarrow \text{ This gives back (-4, 2).}
\]

From this analysis, we see that the reflection of a point \( (x, 0) \) produces a new point \( (x, 2) \), and thus confirms that the other original points are either not possible or just reflect themselves.

Based on our evaluations:
**The correct response is \( Z'(4, 0) \)** since it correctly reflects to one of the valid original coordinate points.