Question

find the image of (1,2) after a reflection about y = -1 followed by a reflection about y = 1. ( ? , ) enter the number that belongs in the green box.

Explanation:

Step1: Reflect over \( y = -1 \)

The formula for reflecting a point \((x, y)\) over the line \( y = k \) is \( (x, 2k - y) \). For \( y = -1 \) and the point \((1, 2)\), we calculate the new \( y \)-coordinate: \( 2(-1) - 2 = -2 - 2 = -4 \). So the image after first reflection is \((1, -4)\).

Step2: Reflect over \( y = 1 \)

Using the same reflection formula with \( k = 1 \) and the point \((1, -4)\), the new \( y \)-coordinate is \( 2(1) - (-4) = 2 + 4 = 6 \). Wait, no, wait—wait, maybe I made a mistake. Wait, no, let's re - check. Wait, the first reflection: the distance from \( y = 2 \) to \( y=-1 \) is \( 2 - (-1)=3 \) units. So reflecting over \( y = -1 \), we move 3 units below \( y=-1 \), so \( y=-1 - 3=-4 \), so the point is \((1, -4)\). Then reflecting over \( y = 1 \): the distance from \( y=-4 \) to \( y = 1 \) is \( 1-(-4)=5 \) units. So we move 5 units above \( y = 1 \), so \( y=1 + 5 = 6 \)? Wait, no, that can't be. Wait, maybe a better way: the reflection over \( y = k \) is \( (x, 2k - y) \). So first reflection: \( k=-1 \), so \( (1, 2(-1)-2)=(1, -4) \). Second reflection: \( k = 1 \), so \( (1, 2(1)-(-4))=(1, 2 + 4)=(1, 6) \)? But the green box is for the \( x \)-coordinate? Wait, the original point has \( x = 1 \), and reflections over horizontal lines ( \( y = k \)) do not change the \( x \)-coordinate. So the \( x \)-coordinate remains 1.

Answer:

1