Question

write the coordinates of the vertices after a reflection over the line x = -3. graph with a coordinate grid, points k, l, m, n (forming a rectangle), the line x = -3 (blue vertical line), and input boxes for k, l, m, n coordinates

Explanation:

First, we need to find the original coordinates of the vertices \( K \), \( L \), \( M \), and \( N \) from the graph.

- From the graph, the coordinates of \( K \) are \( (-1, 0) \) (wait, no, looking at the grid, let's recheck. Wait, the blue line is \( x = -3 \). Let's find the original coordinates:

Looking at the grid:

- \( K \): Let's see, the x-coordinate: from the grid, \( K \) is at \( x = -1 \)? Wait, no, the grid lines: the vertical line \( x = -3 \) is blue. Let's check the original points:

Wait, the yellow figure: \( K \) is at \( ( -1, 0) \)? Wait, no, let's count the grid squares. Wait, the x-axis: from -10 to 10, each grid is 1 unit. Let's see:

- \( K \): x = -1, y = 0? Wait, no, looking at the graph, \( K \) is at ( -1, 0)? Wait, no, the blue line is \( x = -3 \). Let's find the original coordinates:

Wait, the original points:

- \( K \): Let's see, the x-coordinate: from the grid, \( K \) is at \( x = -1 \)? Wait, no, maybe I made a mistake. Wait, the figure is a rectangle. Let's list the original coordinates:

Looking at the graph:

- \( K \): ( -1, 0)
- \( L \): ( -1, 4)
- \( M \): ( 2, 4)
- \( N \): ( 2, 0)

Wait, no, wait, the blue line is \( x = -3 \). To reflect over \( x = -3 \), we use the formula for reflection over a vertical line \( x = a \): the new x-coordinate is \( 2a - x \), where \( x \) is the original x-coordinate, and the y-coordinate remains the same.

So the formula for reflection over \( x = a \) is \( (x, y) ightarrow (2a - x, y) \). Here, \( a = -3 \).

So let's find the original coordinates correctly:

Looking at the graph:

- \( K \): Let's see, the x-coordinate: from the grid, \( K \) is at \( x = -1 \)? Wait, no, the grid lines: the vertical line \( x = -3 \) is blue. Let's check the original points:

Wait, the yellow figure: \( K \) is at ( -1, 0)? Wait, no, maybe the original coordinates are:

Wait, the x-axis: each grid is 1 unit. Let's count:

- \( K \): x = -1, y = 0
- \( L \): x = -1, y = 4
- \( M \): x = 2, y = 4
- \( N \): x = 2, y = 0

Wait, let's confirm:

From the graph, \( K \) is at ( -1, 0), \( L \) is at ( -1, 4), \( M \) is at ( 2, 4), \( N \) is at ( 2, 0).

Now, reflecting over \( x = -3 \). The formula for reflection over \( x = a \) is \( (x, y) ightarrow (2a - x, y) \). So \( a = -3 \), so the new x-coordinate is \( 2(-3) - x = -6 - x \), y-coordinate remains \( y \).

Let's compute for each point:

Step 1: Reflect \( K(-1, 0) \)

New x-coordinate: \( -6 - (-1) = -6 + 1 = -5 \)
y-coordinate: 0
So \( K'(-5, 0) \)

Step 2: Reflect \( L(-1, 4) \)

New x-coordinate: \( -6 - (-1) = -5 \)
y-coordinate: 4
So \( L'(-5, 4) \)

Step 3: Reflect \( M(2, 4) \)

New x-coordinate: \( -6 - 2 = -8 \)
y-coordinate: 4
So \( M'(-8, 4) \)

Step 4: Reflect \( N(2, 0) \)

New x-coordinate: \( -6 - 2 = -8 \)
y-coordinate: 0
So \( N'(-8, 0) \)

Wait, but let's check the original coordinates again. Maybe I misread the original points. Let's re-examine the graph:

Looking at the grid, the blue line is \( x = -3 \). The original points:

- \( K \): Let's see, the x-coordinate: from the grid, \( K \) is at \( x = -1 \)? Wait, no, the grid lines: the vertical line \( x = -3 \) is blue. Let's count the distance from \( x = -3 \) to \( K \).

Wait, maybe the original coordinates are:

- \( K \): ( -1, 0) – distance from \( x = -3 \) is \( | -1 - (-3) | = 2 \) units to the right. So reflection should be 2 units to the left of \( x = -3 \), so \( x = -3 - 2 = -5 \), which matches the earlier calculation.

- \( L \): ( -1, 4) – same x as \( K \), so reflection x is -5, y=4.

- \( M \): ( 2, 4) – distance from \( x = -3 \) is \( | 2 - (-3) | = 5 \) units to the right. So reflection x is \( -3 - 5 = -8 \), y=4.

- \( N \): ( 2, 0) – same x as \( M \), so reflection x is -8, y=0.

Yes, that makes sense.

Answer:

\( K'(-5, 0) \)

\( L'(-5, 4) \)

\( M'(-8, 4) \)

\( N'(-8, 0) \)