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 point \((1, 2)\), we calculate the new \( y \)-coordinate: \( 2(-1) - 2 = -2 - 2 = -4 \)? Wait, no, wait. Wait, the distance from \( y = 2 \) to \( y = -1 \) is \( 2 - (-1) = 3 \) units. So reflecting over \( y = -1 \), we move 3 units below \( y = -1 \)? Wait, no, reflection: the distance from the point to the line is the same on both sides. So the \( y \)-coordinate of the point is \( 2 \), the line is \( y = -1 \). The distance between \( 2 \) and \( -1 \) is \( |2 - (-1)| = 3 \). So the reflected point over \( y = -1 \) will be \( 3 \) units below \( y = -1 \)? Wait, no: if the point is above the line, the reflection is below. So the new \( y \)-coordinate is \( -1 - 3 = -4 \)? Wait, no, let's use the formula. The formula for reflection over \( y = k \) is \( (x, 2k - y) \). So \( k = -1 \), \( y = 2 \), so \( 2(-1) - 2 = -2 - 2 = -4 \). So after first reflection, the point is \( (1, -4) \).

Step2: Reflect over \( y = 1 \)

Now we reflect \( (1, -4) \) over \( y = 1 \). Using the same formula, \( k = 1 \), \( y = -4 \). So new \( y \)-coordinate is \( 2(1) - (-4) = 2 + 4 = 6 \)? Wait, that can't be right. Wait, maybe I made a mistake in the first step. Wait, let's re-examine the first reflection. The original point is \( (1, 2) \), line \( y = -1 \). The vertical distance from \( (1, 2) \) to \( y = -1 \) is \( 2 - (-1) = 3 \). So to reflect, we go 3 units below \( y = -1 \), so \( -1 - 3 = -4 \). That seems correct. Then reflecting \( (1, -4) \) over \( y = 1 \). The vertical distance from \( -4 \) to \( 1 \) is \( 1 - (-4) = 5 \). So we go 5 units above \( y = 1 \), so \( 1 + 5 = 6 \). Wait, but that seems off. Wait, maybe there's a better way. Alternatively, two reflections over parallel lines (both horizontal, so parallel) is equivalent to a translation. The distance between \( y = -1 \) and \( y = 1 \) is \( 2 \) units. So reflecting over \( y = -1 \) then \( y = 1 \) is a translation of \( 2 \times 2 = 4 \) units up? Wait, original point \( (1, 2) \). Let's check the vertical distance from \( 2 \) to \( -1 \) is 3, then from \( -1 \) to \( 1 \) is 2, so total? Wait, no, maybe the formula for two reflections over \( y = k_1 \) and \( y = k_2 \) (parallel) is a translation of \( 2(k_2 - k_1) \) in the y-direction. So \( k_1 = -1 \), \( k_2 = 1 \), so \( 2(1 - (-1)) = 4 \). So the original \( y \)-coordinate is 2, so new \( y \)-coordinate is \( 2 + 4 = 6 \)? Wait, but let's do it step by step again.

First reflection: \( (1, 2) \) over \( y = -1 \). The midpoint between \( y = 2 \) and the reflected \( y \)-coordinate should be \( y = -1 \). Let the reflected \( y \)-coordinate be \( y_1 \). Then \( \frac{2 + y_1}{2} = -1 \). Solving: \( 2 + y_1 = -2 \) → \( y_1 = -4 \). So first reflection: \( (1, -4) \).

Second reflection: \( (1, -4) \) over \( y = 1 \). Let the reflected \( y \)-coordinate be \( y_2 \). Then \( \frac{-4 + y_2}{2} = 1 \). Solving: \( -4 + y_2 = 2 \) → \( y_2 = 6 \). So the final point is \( (1, 6) \). Wait, but that seems high. Wait, maybe I messed up the direction. Wait, the original point is \( (1, 2) \), which is above \( y = 1 \) (since \( 2 > 1 \)) and above \( y = -1 \). Reflecting over \( y = -1 \) first: moving down to \( y = -4 \), then reflecting over \( y = 1 \): moving up from \( y = -4 \) to \( y = 6 \). Alternatively, maybe the problem is that the two reflections are over \( y = -1 \) then \( y = 1 \), so the total vertical shift is \( 2*(1 - (-1)) = 4 \), so original \( y = 2 \), so \( 2 + 4 = 6 \). So the green box should be 6.

Wait, but let's check with the grid. The original point is at (1,2). Reflect over y=-1: the distance from 2 to -1 is 3, so below y=-1 by 3 is -4, so (1,-4). Then reflect over y=1: distance from -4 to 1 is 5, so above y=1 by 5 is 6. So yes, (1,6). So the answer is 6.

Answer:

6