Question

Use the image to answer the question.

Triangle upper X upper Y upper Z and a dotted line are graphed on a coordinate plane. The x- and y-axes range from negative 7 to 7 in increments of 1. The dotted line is labeled axis of reflection and follows y equals 1.

What are the coordinates of Y′
after a reflection across y=1
?

(1 point)
Responses

Y′(4,−1)
upper Y prime left parenthesis 4 comma negative 1 right parenthesis

Y′(−2,−1)
upper Y prime left parenthesis negative 2 comma negative 1 right parenthesis

Y′(−2,3)
upper Y prime left parenthesis negative 2 comma 3 right parenthesis

Y′(4,5)

Answer 1

To find the coordinates of \( Y' \) after a reflection across the line \( y = 1 \), we need to determine the original coordinates of point \( Y \).

Assuming point \( Y \) has coordinates \( (x, y) \):

1. The distance from point \( Y \) to the line \( y = 1 \) is calculated as \( y - 1 \).
2. The reflection \( Y' \) will be the same distance above the line \( y = 1 \). Therefore, we can find \( Y' \) by calculating \( Y' = 1 - (y - 1) = 2 - y \).

Let's verify which of the provided options matches this transformation:

- If \( Y(4, -1) \):
- Distance to \( y = 1 \): \(-1 - 1 = -2\)
- Reflection: \( 1 - (-2) = 3\) => Y' is at \( (4, 3) \). Not an option.

- If \( Y(-2, -1) \):
- Distance to \( y = 1 \): \(-1 - 1 = -2\)
- Reflection: \( 1 - (-2) = 3\) => Y' is at \( (-2, 3) \). This matches.

- If \( Y(-2, 3) \):
- Distance to \( y = 1 \): \(3 - 1 = 2\)
- Reflection: \( 1 - 2 = -1\) => Y' is at \( (-2, -1) \). Not an option.

- If \( Y(4, 5) \):
- Distance to \( y = 1 \): \(5 - 1 = 4\)
- Reflection: \( 1 - 4 = -3\) => Y' is at \( (4, -3) \). Not an option.

Thus, the correct answer based on the reflection across \( y = 1 \) is:

**Y′(−2,3)**