Question

write the coordinates of the vertices after a reflection over the line x = -1. graph with points q, r, s, t on a coordinate grid and input boxes for q, r, s, t coordinates

Explanation:

Step1: Find original coordinates

First, identify the original coordinates of each vertex from the graph:
- \( Q(-9, -2) \)
- \( R(-9, 1) \)
- \( S(-1, 3) \)
- \( T(-1, 0) \)

Step2: Reflection over \( x = -1 \)

The formula for reflecting a point \( (x, y) \) over the vertical line \( x = a \) is \( (2a - x, y) \). Here, \( a = -1 \), so the formula becomes \( (2(-1) - x, y) = (-2 - x, y) \).

For \( Q(-9, -2) \):

Apply the formula: \( x' = -2 - (-9) = -2 + 9 = 7 \), \( y' = -2 \). So \( Q'(7, -2) \)? Wait, no, wait. Wait, let's check the distance from \( x = -1 \). The original \( x \)-coordinate of \( Q \) is \( -9 \). The distance between \( -9 \) and \( -1 \) is \( |-9 - (-1)| = |-8| = 8 \). So the reflected \( x \)-coordinate should be \( -1 + 8 = 7 \)? Wait, no, wait: if the point is to the left of \( x = -1 \), the reflection is to the right, at the same distance. Wait, \( x = -1 \) is the line. So for a point \( (x, y) \), the reflection over \( x = a \) is \( (2a - x, y) \). So \( a = -1 \), so \( 2*(-1) - x = -2 - x \). Let's recalculate:

For \( Q(-9, -2) \):
\( x' = -2 - (-9) = -2 + 9 = 7 \), \( y' = -2 \). So \( Q'(7, -2) \)? Wait, but let's check the graph. Wait, maybe I made a mistake. Wait, the original coordinates: let's look at the graph again. The grid: each square is 1 unit. Let's find the coordinates:

- \( Q \): looks like \( x = -9 \), \( y = -2 \) (since it's 9 units left of origin, 2 units down)
- \( R \): \( x = -9 \), \( y = 1 \) (9 left, 1 up)
- \( S \): \( x = -1 \), \( y = 3 \) (1 left, 3 up)
- \( T \): \( x = -1 \), \( y = 0 \) (1 left, on x-axis)

Now, reflection over \( x = -1 \). The line \( x = -1 \) is a vertical line. For a vertical line reflection, the \( y \)-coordinate remains the same, and the \( x \)-coordinate is reflected across \( x = -1 \).

The distance from a point \( (x, y) \) to the line \( x = -1 \) is \( |x - (-1)| = |x + 1| \). So the reflected \( x \)-coordinate is \( -1 + ( - (x + 1)) = -1 - x - 1 = -x - 2 \)? Wait, no: if the point is at \( x \), the reflection over \( x = a \) is \( 2a - x \). So \( a = -1 \), so \( 2*(-1) - x = -2 - x \). Let's verify with \( S \): \( S(-1, 3) \). Then \( x' = -2 - (-1) = -2 + 1 = -1 \), which is correct, since \( S \) is on the line \( x = -1 \), so it reflects to itself. Good.

For \( T(-1, 0) \): \( x' = -2 - (-1) = -1 \), \( y' = 0 \), so \( T'(-1, 0) \), correct.

For \( R(-9, 1) \): \( x' = -2 - (-9) = 7 \), \( y' = 1 \). So \( R'(7, 1) \).

For \( Q(-9, -2) \): \( x' = -2 - (-9) = 7 \), \( y' = -2 \). So \( Q'(7, -2) \).

Wait, but let's check the distance. The distance from \( R(-9, 1) \) to \( x = -1 \) is \( |-9 - (-1)| = 8 \). So the reflected point should be 8 units to the right of \( x = -1 \), so \( -1 + 8 = 7 \). Correct. Same for \( Q \): distance is 8, so \( -1 + 8 = 7 \). Correct.

For \( S(-1, 3) \): distance is 0, so reflected point is same: \( S'(-1, 3) \).

For \( T(-1, 0) \): distance is 0, so reflected point is same: \( T'(-1, 0) \).

Wait, but let's confirm the original coordinates again. Maybe I misread the graph. Let's look at the grid:

- The x-axis: from -10 to 10. The y-axis: from -10 to 10.

- Point Q: looks like (x=-9, y=-2) (since it's at x=-9, y=-2)
- Point R: (x=-9, y=1) (x=-9, y=1)
- Point S: (x=-1, y=3) (x=-1, y=3)
- Point T: (x=-1, y=0) (x=-1, y=0)

Yes, that's correct.

So applying the reflection formula \( (2a - x, y) \) where \( a = -1 \):

- \( Q(-9, -2) \): \( x' = 2*(-1) - (-9) = -2 + 9 = 7 \), \( y' = -2 \) → \( Q'(7, -2) \)
- \( R(-9, 1) \): \( x' = 2*(-1) - (-9) = -2 + 9 = 7 \), \( y' = 1 \) → \( R'(7, 1) \)
- \( S(-1, 3) \): \( x' = 2*(-1) - (-1) = -2 + 1 = -1 \), \( y' = 3 \) → \( S'(-1, 3) \)
- \( T(-1, 0) \): \( x' = 2*(-1) - (-1) = -2 + 1 = -1 \), \( y' = 0 \) → \( T'(-1, 0) \)

Wait, but let's check with another method. The reflection over a vertical line \( x = a \) can be thought of as: the midpoint between the original point \( (x, y) \) and the reflected point \( (x', y) \) lies on the line \( x = a \). So \( \frac{x + x'}{2} = a \) → \( x' = 2a - x \). Which is the same formula. So that's correct.

So the coordinates after reflection are:

- \( Q'(7, -2) \)
- \( R'(7, 1) \)
- \( S'(-1, 3) \)
- \( T'(-1, 0) \)

Answer:

\( Q'(7, -2) \), \( R'(7, 1) \), \( S'(-1, 3) \), \( T'(-1, 0) \)