Question

if the figure below has been reflected across the x-axis, what are the coordinates of the preimage?

Explanation:

Step1: Find coordinates of \( P' \), \( Q' \), \( R' \)

From the graph:
- \( P' \) is at \( (3, 4) \)
- \( Q' \) is at \( (-2, 1) \)
- \( R' \) is at \( (1, -5) \) (Wait, correction: Looking at the graph, \( R' \) is at \( (1, -5) \)? Wait, no, the y-axis: the bottom point \( R' \) is at \( (1, -5) \)? Wait, actually, let's recheck. The grid: x-axis and y-axis. Let's see \( P' \): x=3, y=4 (since it's 3 units right on x, 4 up on y). \( Q' \): x=-2, y=1 (2 left on x, 1 up on y). \( R' \): x=1, y=-5? Wait, no, the bottom point: looking at the graph, the point \( R' \) is at (1, -5)? Wait, maybe I misread. Wait, the reflection over x-axis: the rule is \( (x, y) \) reflects to \( (x, -y) \) for preimage to image? Wait, no: if the image is after reflection over x-axis, then preimage \( (x, y) \) reflects to image \( (x, -y) \). Wait, no: reflection over x-axis: the image of a point \( (a, b) \) is \( (a, -b) \). So to find preimage, if image is \( (a, b) \), preimage is \( (a, -b) \)? Wait, no: Wait, reflection over x-axis: the preimage (original) point \( (x, y) \) becomes image \( (x, -y) \). So if we have the image, to find preimage, we need to reverse: if image is \( (x', y') \), then preimage is \( (x', -y') \)? Wait, no: Let's clarify.

Reflection over x-axis: The x-coordinate remains the same, the y-coordinate is multiplied by -1. So if preimage is \( (x, y) \), image is \( (x, -y) \). Therefore, to find preimage from image, we take image \( (x', y') \) and preimage is \( (x', -y') \)? Wait, no: Wait, suppose preimage is \( (x, y) \), after reflection over x-axis, image is \( (x, -y) \). So if we know the image \( (x', y') \), then \( x' = x \), \( y' = -y \), so \( y = -y' \). Therefore, preimage is \( (x', -y') \).

So let's find coordinates of \( P' \), \( Q' \), \( R' \):

- \( P' \): From graph, x=3, y=4. So image \( P' = (3, 4) \). Therefore, preimage \( P = (3, -4) \)? Wait, no: Wait, if preimage is reflected over x-axis to get image, then preimage \( (x, y) \) → image \( (x, -y) \). So if image is \( (3, 4) \), then preimage \( (x, y) \) must satisfy \( x = 3 \), \( -y = 4 \) → \( y = -4 \). So preimage \( P = (3, -4) \)? Wait, that seems wrong. Wait, maybe I mixed up preimage and image. The problem says: "the figure below has been reflected across the x-axis, what are the coordinates of the preimage?" So the figure shown is the image (after reflection), and we need to find the preimage (before reflection). So reflection over x-axis: preimage → image: \( (x, y) \) → \( (x, -y) \). Therefore, image → preimage: \( (x, y) \) → \( (x, -y) \)? Wait, no: If you reflect over x-axis, the preimage (original) is reflected to get the image. So to get preimage from image, you reverse the reflection. So reflection over x-axis: (x, y) → (x, -y). So inverse: (x, y) → (x, -y) as well? Wait, no: reflection over x-axis is its own inverse? Wait, no: If you reflect a point over x-axis, then reflect it again over x-axis, you get back the original. So the inverse of reflection over x-axis is the same reflection. So if image is \( (x, y) \), preimage is \( (x, -y) \)? Wait, let's take an example: preimage (2, 3), reflect over x-axis: (2, -3) (image). Now, if we have image (2, -3), preimage is (2, 3), which is (2, -(-3)) = (2, 3). Ah! So yes: to get preimage from image (after reflection over x-axis), we take the image's y-coordinate, multiply by -1. So preimage \( (x, y) \) where image is \( (x, y') \), so \( y = -y' \).

So let's re-express:

- \( P' \): coordinates (3, 4) (image). So preimage \( P \): x=3, y= -4? Wait, no: Wait, image is (3, 4), so preimage is (3, -4)? Wait, no, wait: preimage (x, y) → image (x, -y). So if image is (3, 4), then -y = 4 → y = -4. So preimage P is (3, -4).

- \( Q' \): coordinates (-2, 1) (image). So preimage Q: x=-2, y= -1? Wait, no: -y = 1 → y = -1? Wait, no: image is (x', y') = (-2, 1). So preimage (x, y) has x = x' = -2, and -y = y' = 1 → y = -1. So Q is (-2, -1)? Wait, no, that can't be. Wait, maybe I got the image wrong. Let's look at the graph again.

Looking at the graph:

- \( P' \): is at (3, 4) (x=3, y=4, since it's 3 units right on x, 4 up on y).

- \( Q' \): is at (-2, 1) (x=-2, y=1, 2 units left on x, 1 up on y).

- \( R' \): is at (1, -5)? Wait, no, the bottom point: looking at the graph, the point \( R' \) is at (1, -5)? Wait, the y-axis: the bottom of the triangle is at (1, -5)? Wait, maybe the coordinates are:

Wait, the grid: each square is 1 unit. So:

- \( P' \): (3, 4)

- \( Q' \): (-2, 1)

- \( R' \): (1, -5)

Now, reflection over x-axis: preimage (x, y) → image (x, -y). So to find preimage, we take image (x, y) and preimage is (x, -y). Wait, no: preimage (x, y) → image (x, -y). So image (x, -y) → preimage (x, y). So if image is (3, 4), then preimage is (3, -4)? Wait, no: image is (x, -y) = (3, 4) → -y = 4 → y = -4. So preimage P is (3, -4).

Image \( Q' = (-2, 1) \): so (x, -y) = (-2, 1) → -y = 1 → y = -1. So preimage Q is (-2, -1)? Wait, that doesn't seem right. Wait, maybe the image is the reflected one, so preimage is the original, and image is after reflection. So if the figure shown is the image (after reflection), then preimage is before reflection. So let's take a point: suppose preimage is (3, -4), reflect over x-axis: (3, 4), which is \( P' \). That works.

Preimage Q: suppose preimage is (-2, -1), reflect over x-axis: (-2, 1), which is \( Q' \). That works.

Preimage R: suppose preimage is (1, 5), reflect over x-axis: (1, -5), which is \( R' \). Ah! So here's the mistake: \( R' \) is (1, -5), so preimage R is (1, 5), because reflecting (1, 5) over x-axis gives (1, -5).

So let's correct:

- \( P' \): (3, 4) → preimage P: (3, -4) (since reflecting (3, -4) over x-axis: (3, 4))

- \( Q' \): (-2, 1) → preimage Q: (-2, -1) (reflecting (-2, -1) over x-axis: (-2, 1))

- \( R' \): (1, -5) → preimage R: (1, 5) (reflecting (1, 5) over x-axis: (1, -5))

Wait, that makes sense. Let's verify:

Reflection over x-axis: (x, y) → (x, -y). So:

- P: (3, -4) → reflect: (3, 4) = \( P' \) ✔️

- Q: (-2, -1) → reflect: (-2, 1) = \( Q' \) ✔️

- R: (1, 5) → reflect: (1, -5) = \( R' \) ✔️

Yes, that works. So the coordinates of the preimage points are:

- P: (3, -4)

- Q: (-2, -1)

- R: (1, 5)

Wait, but let's check the graph again. The \( R' \) is at the bottom, y=-5, so preimage R is at y=5 (since reflection over x-axis flips y-sign). So yes.

Step1: Identify image coordinates

- \( P' \): \( (3, 4) \)
- \( Q' \): \( (-2, 1) \)
- \( R' \): \( (1, -5) \)

Step2: Apply reflection over x-axis rule (preimage to image: \( (x, y) \to (x, -y) \); so image to preimage: \( (x, y) \to (x, -y) \) [since reflection is its own inverse? Wait, no: preimage (x,y) → image (x,-y). So image (x,-y) → preimage (x,y). So to get preimage from image (x,y), preimage is (x, -y)? Wait, no: image is (x, -y), so preimage is (x, y). Wait, I think I confused the notation. Let's re-express:

Let preimage point be \( (x, y) \). After reflection over x-axis, the image point is \( (x, -y) \). So if we know the image point \( (x', y') \), then:

\( x' = x \) (x-coordinate same)

\( y' = -y \) (y-coordinate is negative of preimage's y)

Therefore, to find preimage \( (x, y) \) from image \( (x', y') \):

\( x = x' \)

\( y = -y' \)

Ah! There we go. So:

For \( P' = (3, 4) \):

\( x = 3 \), \( y = -4 \) → preimage \( P = (3, -4) \)

For \( Q' = (-2, 1) \):

\( x = -2 \), \( y = -1 \) → preimage \( Q = (-2, -1) \)

For \( R' = (1, -5) \):

\( x = 1 \), \( y = -(-5) = 5 \) → preimage \( R = (1, 5) \)

Yes, that's correct. So the preimage coordinates are:

- P: (3, -4)

- Q: (-2, -1)

- R: (1, 5)

Answer:

• P: \((3, -4)\)
• Q: \((-2, -1)\)
• R: \((1, 5)\)