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URL: https://nakafa.com/en/subjects/mathematics/geometric-transformation/reflection-over-y-equals-minus-x
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Study reflection over y=-x through coordinate-swapping rules and examples.

---

## Understanding Reflection over the Negative Diagonal Line

Reflection over the line $$y = -x$$ is a geometric transformation where each point of an object is mapped to a new position, with the line $$y = -x$$ acting as a mirror.

Visible text: Reflection over the line is a geometric transformation where each point of an object is mapped to a new position, with the line acting as a mirror.

The line connecting the original point to its image will be perpendicular to the line $$y = -x$$, and the distance from the original point to the mirror line is equal to the distance from its image to the mirror line.

Visible text: The line connecting the original point to its image will be perpendicular to the line , and the distance from the original point to the mirror line is equal to the distance from its image to the mirror line.

### Rule for Reflection over the Negative Diagonal Line

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

Visible text: If a point is reflected over the line , its image's coordinates, <InlineMath math="P'(x', y')" />, are determined by the rule:

Component: MathContainer
Children:

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

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

Thus, the image of point $$P(x, y)$$ is <InlineMath math="P'(-y, -x)" />. Notice that the $$x$$ and $$y$$ coordinates swap places and change signs (become their negatives).

Visible text: Thus, the image of point is <InlineMath math="P'(-y, -x)" />. Notice that the and coordinates swap places and change signs (become their negatives).

## Reflecting a Point

Suppose we have point $$D(2, 1)$$. If point $$D$$ is reflected over the line $$y = -x$$, its image, <InlineMath math="D'" />, is:

Visible text: Suppose we have point . If point is reflected over the line , its image, <InlineMath math="D'" />, is:

Component: MathContainer
Children:

```math
x' = -(1) = -1
```

```math
y' = -(2) = -2
```

Thus, the image of point $$D$$ is <InlineMath math="D'(-1, -2)" />.

Visible text: Thus, the image of point is <InlineMath math="D'(-1, -2)" />.

Let's visualize the reflection of several points over the line $$y = -x$$:

Visible text: Let's visualize the reflection of several points over the line :

Component: LineEquation
Props:
- title: Image of Points over the Line $$y=-x$$
  Visible text: Image of Points over the Line
- description: Visualization of the reflection of several points over the line{" "}
$$y=-x$$.
  Visible text: Visualization of the reflection of several points over the line{" "}
.
- data: [
{
points: [
{ x: -6, y: 6, z: 0 },
{ x: 6, y: -6, z: 0 },
],
color: getColor("INDIGO"),
labels: [{ text: "y=-x", at: 1, offset: [0.5, 0.5, 0] }],
}, // Reflection line
// Point A and A'
{
points: [{ x: -2, y: 4, z: 0 }],
color: getColor("AMBER"),
showPoints: true,
labels: [{ text: "A(-2,4)", at: 0, offset: [0.3, 0.3, 0] }],
},
{
points: [{ x: -4, y: 2, z: 0 }],
color: getColor("AMBER"),
showPoints: true,
labels: [{ text: "A'(-4,2)", at: 0, offset: [0.3, 0.3, 0] }],
},
// Point D and D'
{
points: [{ x: 2, y: 1, z: 0 }],
color: getColor("SKY"),
showPoints: true,
labels: [{ text: "D(2,1)", at: 0, offset: [0.3, 0.3, 0] }],
},
{
points: [{ x: -1, y: -2, z: 0 }],
color: getColor("SKY"),
showPoints: true,
labels: [{ text: "D'(-1,-2)", at: 0, offset: [0.3, 0.3, 0] }],
},
]
- showZAxis: false
- cameraPosition: [0, 0, 15]

## Reflecting a Triangle

Determine the image of triangle $$ABC$$ with vertices $$A(-2,4)$$, $$B(3,1)$$, and $$C(-3,-1)$$ reflected over the line $$y=-x$$.

Visible text: Determine the image of triangle with vertices , , and reflected over the line .

To reflect the triangle, we reflect each of its vertices:

1.  Point $$A(-2, 4)$$: Its image is <InlineMath math="A'(-4, 2)" />.
2.  Point $$B(3, 1)$$: Its image is <InlineMath math="B'(-1, -3)" />.
3.  Point $$C(-3, -1)$$: Its image is <InlineMath math="C'(1, 3)" />.

Visible text: 1. Point : Its image is <InlineMath math="A'(-4, 2)" />.
2. Point : Its image is <InlineMath math="B'(-1, -3)" />.
3. Point : Its image is <InlineMath math="C'(1, 3)" />.

The image triangle <InlineMath math="A'B'C'" /> is formed by connecting the points <InlineMath math="A'(-4, 2)" />, <InlineMath math="B'(-1, -3)" />, and <InlineMath math="C'(1, 3)" />.

Component: LineEquation
Props:
- title: Triangle $$ABC$$ and its Image{" "}
<InlineMath math="A'B'C'" /> over $$y=-x$$
  Visible text: Triangle and its Image{" "}
<InlineMath math="A'B'C'" /> over
- description: Visualization of the reflection of triangle $$ABC$$ over
the line $$y=-x$$.
  Visible text: Visualization of the reflection of triangle over
the line .
- data: [
{
points: [
{ x: -7, y: 7, z: 0 },
{ x: 7, y: -7, z: 0 },
],
color: getColor("INDIGO"),
labels: [{ text: "y=-x", at: 1, offset: [0.5, 0.5, 0] }],
}, // Reflection line
// Triangle ABC (Original)
...[
{
from: { x: -2, y: 4, z: 0, label: "A(-2,4)" },
to: { x: 3, y: 1, z: 0, label: "B(3,1)" },
},
{
from: { x: 3, y: 1, z: 0, label: "B(3,1)" },
to: { x: -3, y: -1, z: 0, label: "C(-3,-1)" },
},
{
from: { x: -3, y: -1, z: 0, label: "C(-3,-1)" },
to: { x: -2, y: 4, z: 0, label: "A(-2,4)" },
},
].map((segment) => ({
points: [segment.from, segment.to],
color: getColor("AMBER"),
showPoints: true,
labels: [{ text: segment.from.label, at: 0, offset: [0.4, 0.4, 0] }],
})),
// Triangle A'B'C' (Image)
...[
{
from: { x: -4, y: 2, z: 0, label: "A'(-4,2)" },
to: { x: -1, y: -3, z: 0, label: "B'(-1,-3)" },
},
{
from: { x: -1, y: -3, z: 0, label: "B'(-1,-3)" },
to: { x: 1, y: 3, z: 0, label: "C'(1,3)" },
},
{
from: { x: 1, y: 3, z: 0, label: "C'(1,3)" },
to: { x: -4, y: 2, z: 0, label: "A'(-4,2)" },
},
].map((segment) => ({
points: [segment.from, segment.to],
color: getColor("TEAL"),
showPoints: true,
labels: [{ text: segment.from.label, at: 0, offset: [0.4, 0.4, 0] }],
})),
]
- showZAxis: false
- cameraPosition: [0, 0, 18]

## Reflecting a Line Equation

If a line has the equation $$y = -4x - 2$$ is reflected over the line $$y=-x$$, determine the equation of its image.

Visible text: If a line has the equation is reflected over the line , determine the equation of its image.

To find the equation of the image, we use the rule <InlineMath math="x' = -y" /> and <InlineMath math="y' = -x" />. This means we replace every $$x$$ in the original equation with $$-y$$ and every $$y$$ with $$-x$$.

Visible text: To find the equation of the image, we use the rule <InlineMath math="x' = -y" /> and <InlineMath math="y' = -x" />. This means we replace every in the original equation with and every with .

Original equation:

```math
y = -4x - 2
```

Substitute $$x \rightarrow -y$$ and $$y \rightarrow -x$$:

Visible text: Substitute and :

```math
(-x) = -4(-y) - 2
```

Simplify the equation for the image line:

Component: MathContainer
Children:

```math
-x = 4y - 2
```

```math
-x + 2 = 4y
```

```math
y = -\frac{1}{4}x + \frac{2}{4}
```

```math
y = -\frac{1}{4}x + \frac{1}{2}
```

So, the equation of the image of the line $$y = -4x - 2$$ after reflection over $$y=-x$$ is $$y = -\frac{1}{4}x + \frac{1}{2}$$.

Visible text: So, the equation of the image of the line after reflection over is .

Component: LineEquation
Props:
- title: Line $$y=-4x-2$$ and its Image over{" "}
$$y=-x$$
  Visible text: Line and its Image over{" "}
- description: Reflection of the line $$y=-4x-2$$ over the line{" "}
$$y=-x$$.
  Visible text: Reflection of the line over the line{" "}
.
- data: [
{
points: [
{ x: -7, y: 7, z: 0 },
{ x: 7, y: -7, z: 0 },
],
color: getColor("INDIGO"),
labels: [{ text: "y=-x", at: 1, offset: [0.5, 0.5, 0] }],
}, // Reflection line
{
// Original Line y = -4x - 2
points: Array.from({ length: 11 }, (_, i) => {
const xVal = (i - 5) * 0.5; // x range to prevent it from being too steep
return { x: xVal, y: -4 * xVal - 2, z: 0 };
}),
color: getColor("PURPLE"),
smooth: true,
labels: [{ text: "y=-4x-2", at: 6, offset: [1, 0.5, 0] }],
},
{
// Image Line y = (-1/4)x + 1/2
points: Array.from({ length: 11 }, (_, i) => {
const xVal = (i - 5) * 2; // x range to prevent it from being too flat
return { x: xVal, y: (-1 / 4) * xVal + 1 / 2, z: 0 };
}),
color: getColor("PINK"),
smooth: true,
labels: [{ text: "y=(-1/4)x+1/2", at: 8, offset: [0.5, -1, 0] }],
},
]
- showZAxis: false
- cameraPosition: [0, 0, 18]

## Exercises

1.  Determine the coordinates of the image of point $$S(5, -9)$$ if it is reflected over the line $$y = -x$$!
2.  Determine the image of triangle $$ABC$$ with vertices $$A(1, -2)$$, $$B(2, 3)$$, and $$C(-2, 0)$$ reflected over the line $$y = -x$$.
3.  If a line has the equation $$4x + 3y - 2 = 0$$ is reflected over the line $$y = -x$$, determine the equation of its image.

Visible text: 1. Determine the coordinates of the image of point if it is reflected over the line !
2. Determine the image of triangle with vertices , , and reflected over the line .
3. If a line has the equation is reflected over the line , determine the equation of its image.

### Key Answers

1.  The image of point $$S(5, -9)$$ is <InlineMath math="S'(9, -5)" />.

    **Explanation:**

    <MathContainer>
      <BlockMath math="x' = -(-9) = 9" />
      <BlockMath math="y' = -(5) = -5" />
    </MathContainer>

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

    - <InlineMath math="A'(2, -1)" /> (from $$A(1, -2)$$)
    - <InlineMath math="B'(-3, -2)" /> (from $$B(2, 3)$$)
    - <InlineMath math="C'(0, 2)" /> (from $$C(-2, 0)$$)

3.  The equation of the image of the line $$4x + 3y - 2 = 0$$ is $$4(-y) + 3(-x) - 2 = 0$$.

    **Explanation:** Substitute $$x \rightarrow -y$$ and $$y \rightarrow -x$$ into the original equation:

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

      
    
    ```math
    -4y - 3x - 2 = 0
    ```

      
    
    ```math
    3x + 4y + 2 = 0
    ```

    </MathContainer>

Visible text: 1. The image of point is <InlineMath math="S'(9, -5)" />.

 **Explanation:**

 <MathContainer>
 <BlockMath math="x' = -(-9) = 9" />
 <BlockMath math="y' = -(5) = -5" />
 </MathContainer>

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

 - <InlineMath math="A'(2, -1)" /> (from )
 - <InlineMath math="B'(-3, -2)" /> (from )
 - <InlineMath math="C'(0, 2)" /> (from )

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

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

 <MathContainer>
 
 

 
 

 
 

 </MathContainer>