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URL: https://nakafa.com/en/subjects/mathematics/geometric-transformation/reflection-over-y-axis
Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/material/lesson/mathematics/geometric-transformation/reflection-over-y-axis/en.mdx

Learn y-axis reflection with examples for points, triangles, and lines. Use the P'(-x, y) rule with interactive visualizations.

---

## Understanding Reflection over the Vertical Axis

Reflection over the $$y$$-axis is a type of geometric transformation that moves every point on an object to a new position. Imagine the $$y$$-axis as a mirror. Every point will have an image on the opposite side of the $$y$$-axis at the same distance from the $$y$$-axis.

Visible text: Reflection over the -axis is a type of geometric transformation that moves every point on an object to a new position. Imagine the -axis as a mirror. Every point will have an image on the opposite side of the -axis at the same distance from the -axis.

### Rule for Reflection over the Vertical Axis

If a point $$P(x, y)$$ is reflected over the $$y$$-axis, its image's coordinates, <InlineMath math="P'(x', y')" />, will follow the rule:

Visible text: If a point is reflected over the -axis, its image's coordinates, <InlineMath math="P'(x', y')" />, will follow the rule:

Component: MathContainer
Children:

```math
x' = -x
```

```math
y' = y
```

Thus, the image of point $$P(x, y)$$ is <InlineMath math="P'(-x, y)" />.

Visible text: Thus, the image of point is <InlineMath math="P'(-x, y)" />.

Note that the value of the $$y$$-coordinate does not change, while the value of the $$x$$-coordinate becomes its opposite (negative if positive, positive if negative).

Visible text: Note that the value of the -coordinate does not change, while the value of the -coordinate becomes its opposite (negative if positive, positive if negative).

## Reflecting a Point

Suppose we have point $$A(3, 4)$$. If point $$A$$ is reflected over the $$y$$-axis, its image, <InlineMath math="A'" />, can be determined as follows:

Visible text: Suppose we have point . If point is reflected over the -axis, its image, <InlineMath math="A'" />, can be determined as follows:

The original $$x$$-coordinate is $$3$$, so <InlineMath math="x' = -3" />.

Visible text: The original -coordinate is , so <InlineMath math="x' = -3" />.

The original $$y$$-coordinate is $$4$$, so <InlineMath math="y' = 4" />.

Visible text: The original -coordinate is , so <InlineMath math="y' = 4" />.

Thus, the image of point $$A$$ is <InlineMath math="A'(-3, 4)" />.

Visible text: Thus, the image of point is <InlineMath math="A'(-3, 4)" />.

Let's visualize this:

Component: LineEquation
Props:
- title: Point $$A(3,4)$$ and its Image{" "}
<InlineMath math="A'(-3,4)" />
  Visible text: Point and its Image{" "}
<InlineMath math="A'(-3,4)" />
- description: Visualization of the reflection of point $$A$$ over the{" "}
$$y$$-axis.
  Visible text: Visualization of the reflection of point over the{" "}
-axis.
- data: [
{
points: [{ x: 3, y: 4, z: 0 }],
color: getColor("EMERALD"),
showPoints: true,
labels: [{ text: "A(3,4)", at: 0, offset: [0.3, 0.3, 0] }],
},
{
points: [{ x: -3, y: 4, z: 0 }],
color: getColor("SKY"),
showPoints: true,
labels: [{ text: "A'(-3,4)", at: 0, offset: [0.3, 0.3, 0] }],
},
]
- showZAxis: false
- cameraPosition: [0, 0, 12]

## Reflecting a Triangle

Now, let's reflect a triangle $$PQR$$ with vertices $$P(1, 2)$$, $$Q(4, 4)$$, and $$R(2, 0)$$ over the $$y$$-axis.

Visible text: Now, let's reflect a triangle with vertices , , and over the -axis.

To reflect a triangle, we need to reflect each of its vertices.

1.  Point $$P(1, 2)$$: Its image is <InlineMath math="P'(-1, 2)" />.
2.  Point $$Q(4, 4)$$: Its image is <InlineMath math="Q'(-4, 4)" />.
3.  Point $$R(2, 0)$$: Its image is <InlineMath math="R'(-2, 0)" />.

Visible text: 1. Point : Its image is <InlineMath math="P'(-1, 2)" />.
2. Point : Its image is <InlineMath math="Q'(-4, 4)" />.
3. Point : Its image is <InlineMath math="R'(-2, 0)" />.

By connecting the image points <InlineMath math="P'Q'R'" />, we obtain the reflected triangle.

Component: LineEquation
Props:
- title: Triangle $$PQR$$ and its Image{" "}
<InlineMath math="P'Q'R'" />
  Visible text: Triangle and its Image{" "}
<InlineMath math="P'Q'R'" />
- description: Visualization of the reflection of triangle $$PQR$$ over
the $$y$$-axis.
  Visible text: Visualization of the reflection of triangle over
the -axis.
- data: [
...[
{
from: { x: 1, y: 2, z: 0, label: "P(1,2)" },
to: { x: 4, y: 4, z: 0, label: "Q(4,4)" },
},
{
from: { x: 4, y: 4, z: 0, label: "Q(4,4)" },
to: { x: 2, y: 0, z: 0, label: "R(2,0)" },
},
{
from: { x: 2, y: 0, z: 0, label: "R(2,0)" },
to: { x: 1, y: 2, z: 0, label: "P(1,2)" },
},
].map((segment) => ({
points: [segment.from, segment.to],
color: getColor("AMBER"),
showPoints: true,
labels: [
{ text: segment.from.label, at: 0, offset: [0.5, 0.5, 0] },
{ text: segment.to.label, at: 1, offset: [0.5, 0.5, 0] },
],
})),
...[
{
from: { x: -1, y: 2, z: 0, label: "P'(-1,2)" },
to: { x: -4, y: 4, z: 0, label: "Q'(-4,4)" },
},
{
from: { x: -4, y: 4, z: 0, label: "Q'(-4,4)" },
to: { x: -2, y: 0, z: 0, label: "R'(-2,0)" },
},
{
from: { x: -2, y: 0, z: 0, label: "R'(-2,0)" },
to: { x: -1, y: 2, z: 0, label: "P'(-1,2)" },
},
].map((segment) => ({
points: [segment.from, segment.to],
color: getColor("TEAL"),
showPoints: true,
labels: [
{ text: segment.from.label, at: 0, offset: [-0.5, 0.5, 0] },
{ text: segment.to.label, at: 1, offset: [-0.5, 0.5, 0] },
],
})),
]
- showZAxis: false
- cameraPosition: [0, 0, 12]

## Reflecting a Line Equation

Suppose we have a line with the equation $$y = x + 2$$. To find the equation of its image after reflection over the $$y$$-axis, we substitute $$x$$ with $$-x$$ (because <InlineMath math="x' = -x" />) and $$y$$ with $$y$$ (because <InlineMath math="y' = y" />) into the original equation.

Visible text: Suppose we have a line with the equation . To find the equation of its image after reflection over the -axis, we substitute with (because <InlineMath math="x' = -x" />) and with (because <InlineMath math="y' = y" />) into the original equation.

Original equation:

```math
y = x + 2
```

Substitute $$x \rightarrow -x$$:

Visible text: Substitute :

```math
y = (-x) + 2
```

The equation of the image is:

```math
y = -x + 2
```

Let's visualize these two lines:

Component: LineEquation
Props:
- title: Line $$y = x + 2$$ and its Image{" "}
$$y = -x + 2$$
  Visible text: Line and its Image{" "}
- description: Reflection of a line over the $$y$$-axis.
  Visible text: Reflection of a line over the -axis.
- data: [
{
// Original Line y = x + 2
points: Array.from({ length: 11 }, (_, i) => {
const x = (i - 5) * 0.8; // x from -4 to 4 for better fit
return { x, y: x + 2, z: 0 };
}),
color: getColor("PURPLE"),
smooth: true,
labels: [{ text: "y=x+2", at: 9, offset: [0.7, -0.5, 0] }],
},
{
// Reflected Line y = -x + 2
points: Array.from({ length: 11 }, (_, i) => {
const x = (i - 5) * 0.8; // x from -4 to 4
return { x, y: -x + 2, z: 0 };
}),
color: getColor("PINK"),
smooth: true,
labels: [{ text: "y=-x+2", at: 1, offset: [-0.7, -0.5, 0] }],
},
]
- showZAxis: false
- cameraPosition: [0, 0, 15]

## Exercises

1.  Determine the coordinates of the image of point $$K(-5, 8)$$ if it is reflected over the $$y$$-axis!
2.  A triangle $$ABC$$ has vertices $$A(2, 1)$$, $$B(5, 3)$$, and $$C(3, 6)$$. Determine the coordinates of the image triangle <InlineMath math="A'B'C'" /> after reflection over the $$y$$-axis!
3.  Determine the equation of the image of the line $$3x - 2y + 6 = 0$$ if it is reflected over the $$y$$-axis!
4.  A line has the equation $$y = -3x - 4$$. Determine the equation of its image after reflection over the $$y$$-axis.

Visible text: 1. Determine the coordinates of the image of point if it is reflected over the -axis!
2. A triangle has vertices , , and . Determine the coordinates of the image triangle <InlineMath math="A'B'C'" /> after reflection over the -axis!
3. Determine the equation of the image of the line if it is reflected over the -axis!
4. A line has the equation . Determine the equation of its image after reflection over the -axis.

### Key Answers

1.  The image of point $$K(-5, 8)$$ is <InlineMath math="K'(5, 8)" />.

    **Explanation:** <InlineMath math="x' = -(-5) = 5" />, <InlineMath math="y' = 8" />.

2.  The coordinates of the image triangle <InlineMath math="A'B'C'" /> are:

    - <InlineMath math="A'(-2, 1)" />
    - <InlineMath math="B'(-5, 3)" />
    - <InlineMath math="C'(-3, 6)" />

3.  The equation of the image of the line $$3x - 2y + 6 = 0$$ is $$-3x - 2y + 6 = 0$$.

    **Explanation:** Substitute $$x$$ with $$-x$$ into the original equation:

    <MathContainer>
      
    
    ```math
    3(-x) - 2y + 6 = 0
    ```

      
    
    ```math
    -3x - 2y + 6 = 0
    ```

    </MathContainer>

4.  The equation of the image of the line $$y = -3x - 4$$ is $$y = 3x - 4$$.

    **Explanation:** Substitute $$x$$ with $$-x$$:

    <MathContainer>
      
    
    ```math
    y = -3(-x) - 4
    ```

      
    
    ```math
    y = 3x - 4
    ```

    </MathContainer>

Visible text: 1. The image of point is <InlineMath math="K'(5, 8)" />.

 **Explanation:** <InlineMath math="x' = -(-5) = 5" />, <InlineMath math="y' = 8" />.

2. The coordinates of the image triangle <InlineMath math="A'B'C'" /> are:

 - <InlineMath math="A'(-2, 1)" />
 - <InlineMath math="B'(-5, 3)" />
 - <InlineMath math="C'(-3, 6)" />

3. The equation of the image of the line is .

 **Explanation:** Substitute with into the original equation:

 <MathContainer>
 
 

 
 

 </MathContainer>

4. The equation of the image of the line is .

 **Explanation:** Substitute with :

 <MathContainer>
 
 

 
 

 </MathContainer>