Question

write the coordinates of the vertices after a reflection over the line y = 3.
r(□,□)
s(□,□)
t(□,□)

Explanation:

Step1: Find original coordinates

First, identify the original coordinates of \( R \), \( S \), and \( T \) from the graph.
- \( R \): From the grid, \( R \) is at \( (-5, 5) \) (x = -5, y = 5). Wait, no, looking at the grid: Wait, the y - axis line \( y = 3 \) is the reflection line. Wait, let's re - check the coordinates. Wait, the original points: Let's see, \( R \) is at \( (-5, 5) \)? Wait, no, the grid lines: the x - coordinate of \( R \) is -5? Wait, no, looking at the graph, \( R \) is at \( (-5, 5) \)? Wait, no, the y - coordinate of \( R \) is 5? Wait, the reflection line is \( y = 3 \). Wait, maybe I made a mistake. Let's re - examine: The original points: \( R \): x = -5, y = 5? Wait, no, the grid: each square is 1 unit. Let's see, \( S \) is at \( (0, 5) \)? Wait, no, the graph shows \( S \) at (0,5)? Wait, no, the reflection line is \( y = 3 \). Wait, maybe the original coordinates are: \( R(-5, 5) \), \( S(0, 5) \), \( T(-6, 10) \)? Wait, no, let's look again. Wait, the y - axis (vertical) and x - axis (horizontal). Let's correctly identify:
- \( R \): x = -5, y = 5 (since it's 5 units up from x - axis, x = -5).
- \( S \): x = 0, y = 5 (on the y - axis, y = 5).
- \( T \): x = -6, y = 10 (x = -6, y = 10).

Step2: Reflection over \( y = 3 \)

The formula for reflecting a point \( (x, y) \) over the horizontal line \( y = k \) is \( (x, 2k - y) \). Here, \( k = 3 \), so the new y - coordinate is \( 2\times3 - y=6 - y \), and the x - coordinate remains the same.

For point \( R(-5, 5) \):

New y - coordinate: \( 6 - 5 = 1 \). So \( R'(-5, 1) \).

For point \( S(0, 5) \):

New y - coordinate: \( 6 - 5 = 1 \). So \( S'(0, 1) \).

For point \( T(-6, 10) \):

New y - coordinate: \( 6 - 10=-4 \). So \( T'(-6, -4) \). Wait, that can't be right. Wait, maybe I misidentified the original coordinates. Wait, maybe the original y - coordinates are different. Wait, maybe the original \( R \) is at \( (-5, 5) \)? Wait, no, the reflection line is \( y = 3 \). Let's think again. Maybe the original coordinates are \( R(-5, 5) \), \( S(0, 5) \), \( T(-6, 10) \). Wait, but when we reflect over \( y = 3 \), the distance from the point to the line \( y = 3 \) is \( |y - 3| \), and the reflected point is on the other side of the line at the same distance.

For a point \( (x,y) \), the reflection over \( y = k \) is \( (x, 2k - y) \). So for \( k = 3 \), it's \( (x, 6 - y) \).

Wait, maybe the original coordinates are:
- \( R(-5, 5) \): distance from \( y = 3 \) is \( |5 - 3| = 2 \). So the reflected y - coordinate is \( 3 - 2 = 1 \). So \( R'(-5, 1) \).
- \( S(0, 5) \): distance from \( y = 3 \) is \( |5 - 3| = 2 \). Reflected y - coordinate: \( 3 - 2 = 1 \). So \( S'(0, 1) \).
- \( T(-6, 10) \): distance from \( y = 3 \) is \( |10 - 3| = 7 \). Reflected y - coordinate: \( 3 - 7=-4 \). So \( T'(-6, -4) \). Wait, but that seems odd. Wait, maybe I made a mistake in the original coordinates. Let's re - check the graph. Wait, maybe the original \( R \) is at \( (-5, 5) \), \( S(0, 5) \), \( T(-6, 10) \). Alternatively, maybe the original \( R \) is at \( (-5, 5) \), \( S(0, 5) \), \( T(-6, 10) \).

Wait, another way: Let's look at the y - coordinate of each point. The reflection line is \( y = 3 \). For a point \( (x,y) \), the reflection over \( y = 3 \) is such that the mid - point between \( (x,y) \) and \( (x,y') \) lies on \( y = 3 \). So \( \frac{y + y'}{2}=3\), which gives \( y' = 6 - y \).

Let's re - identify the original coordinates correctly:
- From the graph, \( R \) is at \( (-5, 5) \) (x = -5, y = 5).
- \( S \) is at \( (0, 5) \) (x = 0, y = 5).
- \( T \) is at \( (-6, 10) \) (x = -6, y = 10).

Now, apply the reflection formula \( y'=6 - y \), \( x' = x \):

For \( R(-5, 5) \):

\( x'=-5 \), \( y' = 6 - 5 = 1 \). So \( R'(-5, 1) \).

For \( S(0, 5) \):

\( x' = 0 \), \( y' = 6 - 5 = 1 \). So \( S'(0, 1) \).

For \( T(-6, 10) \):

\( x'=-6 \), \( y' = 6 - 10=-4 \). So \( T'(-6, -4) \).

Answer:

\( R'(-5, 1) \), \( S'(0, 1) \), \( T'(-6, -4) \)