Question

find the image of (1,2) after a reflection about y = -1 followed by a reflection about y = 1. (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 \( (1, 2) \), we calculate \( 2(-1) - 2 = -2 - 2 = -4 \)? Wait, no, wait. Wait, the distance from \( y = 2 \) to \( y = -1 \) is \( 2 - (-1) = 3 \), so the reflection over \( y = -1 \) should be \( -1 - 3 = -4 \)? Wait, no, let's correct. The midpoint between \( y \) and its reflection \( y' \) over \( y = k \) is \( k \). So \( \frac{y + y'}{2} = k \), so \( y' = 2k - y \). So for \( k = -1 \), \( y' = 2(-1) - 2 = -2 - 2 = -4 \)? Wait, but let's check the graph. The point \( (1,2) \), the line \( y = -1 \). The vertical distance from \( y = 2 \) to \( y = -1 \) is \( 2 - (-1) = 3 \) units. So reflecting over \( y = -1 \), we go 3 units below \( y = -1 \), so \( -1 - 3 = -4 \). So after first reflection, the point is \( (1, -4) \).

Step2: Reflect over \( y = 1 \)

Now, reflect \( (1, -4) \) over \( y = 1 \). Using the formula \( y' = 2(1) - (-4) = 2 + 4 = 6 \)? Wait, no, wait. Wait, the midpoint between \( y = -4 \) and \( y' \) should be \( 1 \). So \( \frac{-4 + y'}{2} = 1 \), so \( -4 + y' = 2 \), so \( y' = 6 \)? Wait, that can't be right. Wait, maybe I made a mistake in the first reflection. Wait, let's re-examine the first reflection. The point is \( (1,2) \), line \( y = -1 \). The distance from \( y = 2 \) to \( y = -1 \) is \( 2 - (-1) = 3 \), so the reflection should be \( -1 - 3 = -4 \)? Wait, but let's check with the graph. The line \( y = -1 \) is below \( y = 1 \). Wait, maybe I messed up the formula. Wait, the formula for reflection over horizontal line \( y = k \) is \( (x, 2k - y) \). So for \( (x, y) = (1, 2) \) and \( k = -1 \), \( 2k - y = 2(-1) - 2 = -2 - 2 = -4 \). So first reflection: \( (1, -4) \). Then reflect over \( y = 1 \): \( 2(1) - (-4) = 2 + 4 = 6 \)? But that seems too high. Wait, maybe I made a mistake in the first step. Wait, let's think again. The original point is \( (1,2) \). Reflect over \( y = -1 \). The vertical distance from \( y = 2 \) to \( y = -1 \) is \( 2 - (-1) = 3 \), so the reflection is 3 units below \( y = -1 \), so \( y = -1 - 3 = -4 \). Then reflect over \( y = 1 \). The vertical distance from \( y = -4 \) to \( y = 1 \) is \( 1 - (-4) = 5 \), so the reflection is 5 units above \( y = 1 \), so \( y = 1 + 5 = 6 \). Wait, but that seems odd. Wait, maybe the problem is that the two reflections are over parallel lines, so the composition is a translation. The distance between \( y = -1 \) and \( y = 1 \) is \( 2 \) units. So reflecting over two parallel lines with distance \( d \) results in a translation of \( 2d \) units. The distance between \( y = -1 \) and \( y = 1 \) is \( 2 \), so the total translation is \( 4 \) units? Wait, no. Wait, the first reflection over \( y = -1 \), then over \( y = 1 \). The vector of translation is \( 2(1 - (-1)) = 4 \) units up? Wait, let's check with the original point \( (1,2) \). The first reflection: \( (1, -4) \), then second reflection: \( (1, 6) \). But let's check with the midpoint. Wait, maybe I made a mistake in the first reflection. Wait, let's take a simpler example. Suppose we have a point \( (x, y) \) and reflect over \( y = a \) then \( y = b \), where \( a < b \). The total translation is \( 2(b - a) \). So here, \( a = -1 \), \( b = 1 \), so \( b - a = 2 \), so translation is \( 4 \) units up. So original \( y \)-coordinate is \( 2 \), so \( 2 + 4 = 6 \). Wait, but that seems correct? Wait, but let's check with the graph. The original point is \( (1,2) \), above \( y = 1 \). Reflect over \( y = -1 \), which is below \( y = -1 \), then reflect over \( y = 1 \), which would be above \( y = 1 \). Wait, maybe the answer is 6? But that seems high. Wait, maybe I messed up the formula. Wait, let's rederive the reflection formula. For a horizontal line \( y = k \), the reflection of \( (x, y) \) is \( (x, 2k - y) \). So first reflection over \( y = -1 \): \( (1, 2(-1) - 2) = (1, -2 - 2) = (1, -4) \). Then reflection over \( y = 1 \): \( (1, 2(1) - (-4)) = (1, 2 + 4) = (1, 6) \). So the \( y \)-coordinate is 6.

Answer:

6