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URL: https://nakafa.com/en/subjects/mathematics/geometric-transformation/reflection-over-y-equals-h
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Understand horizontal line reflections with worked examples, visual guides, and point-image rules for geometry problems.

---

## Understanding Reflection over a Horizontal Line

Reflection over the horizontal line $$y = h$$ is a geometric transformation that maps each point of an object to a new position. The line $$y = h$$ acts as a mirror.

Visible text: Reflection over the horizontal line is a geometric transformation that maps each point of an object to a new position. The line acts as a mirror.

The vertical distance from the original point to the mirror line is equal to the vertical distance from the image point to the mirror line. The $$x$$-coordinate of the point does not change.

Visible text: The vertical distance from the original point to the mirror line is equal to the vertical distance from the image point to the mirror line. The -coordinate of the point does not change.

### Rule for Reflection over a Horizontal Line

If a point $$P(x, y)$$ is reflected over the line $$y = h$$, 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' = x
```

```math
y' = 2h - y
```

Thus, the image of point $$P(x, y)$$ is <InlineMath math="P'(x, 2h - y)" />. Note that the $$x$$-coordinate remains the same, while the $$y$$-coordinate changes based on its distance from the line $$y=h$$.

Visible text: Thus, the image of point is <InlineMath math="P'(x, 2h - y)" />. Note that the -coordinate remains the same, while the -coordinate changes based on its distance from the line .

## Reflecting a Point over a Horizontal Line

Determine the image of point $$P(3,2)$$ by reflection over the line $$y=3$$.

Visible text: Determine the image of point by reflection over the line .

In this case, $$x=3$$, $$y=2$$, and $$h=3$$.

Visible text: In this case, , , and .

Using the rule <InlineMath math="P'(x, 2h - y)" />:

Component: MathContainer
Children:

```math
x' = 3
```

```math
y' = 2(3) - 2 = 6 - 2 = 4
```

Thus, the image of point $$P(3,2)$$ is <InlineMath math="P'(3,4)" />.

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

Now, let's visualize this example.

Component: LineEquation
Props:
- title: Image of Point $$P(3,2)$$ over Line{" "}
$$y=3$$
  Visible text: Image of Point over Line{" "}
- description: Visualization of the reflection of point $$P(3,2)$$ over
the line $$y=3$$ resulting in{" "}
<InlineMath math="P'(3,4)" />.
  Visible text: Visualization of the reflection of point over
the line resulting in{" "}
<InlineMath math="P'(3,4)" />.
- data: [
{
points: [
{ x: -5, y: 3, z: 0 },
{ x: 5, y: 3, z: 0 },
],
color: getColor("INDIGO"),
labels: [{ text: "y=3", at: 1, offset: [0.5, 0.5, 0] }],
}, // Reflection line
{
points: [{ x: 3, y: 2, z: 0 }],
color: getColor("EMERALD"),
showPoints: true,
labels: [{ text: "P(3,2)", at: 0, offset: [0.3, -0.5, 0] }],
},
{
points: [{ x: 3, y: 4, z: 0 }],
color: getColor("EMERALD"),
showPoints: true,
labels: [{ text: "P'(3,4)", at: 0, offset: [0.2, 0.3, 0] }],
},
{
points: [
{ x: 3, y: 2, z: 0 },
{ x: 3, y: 4, z: 0 },
],
color: getColor("PINK"),
}, // Vertical helper line
]
- showZAxis: false

## Exercises

1.  Determine the image of point $$P(5,-4)$$ by reflection over the line $$y=-2$$.
2.  A point $$Q(-1, -3)$$ is reflected over the line $$y=0$$ ($$x$$-axis). Determine the coordinates of its image!
3.  The image of a point $$R(x,y)$$ after reflection over the line $$y=4$$ is <InlineMath math="R'(-2,5)" />. Determine the coordinates of point $$R$$!

Visible text: 1. Determine the image of point by reflection over the line .
2. A point is reflected over the line (-axis). Determine the coordinates of its image!
3. The image of a point after reflection over the line is <InlineMath math="R'(-2,5)" />. Determine the coordinates of point !

### Key Answers

1.  Given $$P(5,-4)$$ and the mirror line $$y=-2$$. So $$x=5, y=-4, h=-2$$.

    <MathContainer>
      <BlockMath math="x' = x = 5" />
      <BlockMath math="y' = 2h - y = 2(-2) - (-4) = -4 + 4 = 0" />
    </MathContainer>

    Thus, the image of point $$P$$ is <InlineMath math="P'(5,0)" />.

2.  Given $$Q(-1, -3)$$ and the mirror line $$y=0$$. So $$x=-1, y=-3, h=0$$.

    <MathContainer>
      <BlockMath math="x' = x = -1" />
      <BlockMath math="y' = 2h - y = 2(0) - (-3) = 0 + 3 = 3" />
    </MathContainer>

    Thus, the image of point $$Q$$ is <InlineMath math="Q'(-1,3)" />.

3.  Given the image <InlineMath math="R'(-2,5)" /> and the mirror line $$y=4$$. So <InlineMath math="x'=-2, y'=5, h=4" />.

    We know <InlineMath math="x' = x" /> and <InlineMath math="y' = 2h - y" />.

    From <InlineMath math="x' = x" />, then $$x = -2$$.

    From <InlineMath math="y' = 2h - y" />, then $$5 = 2(4) - y$$.

    <MathContainer>
      
    
    ```math
    5 = 8 - y
    ```

      
    
    ```math
    y = 8 - 5
    ```

      
    
    ```math
    y = 3
    ```

    </MathContainer>

    Thus, the coordinates of point $$R$$ are $$R(-2,3)$$.

Visible text: 1. Given and the mirror line . So .

 <MathContainer>
 <BlockMath math="x' = x = 5" />
 <BlockMath math="y' = 2h - y = 2(-2) - (-4) = -4 + 4 = 0" />
 </MathContainer>

 Thus, the image of point is <InlineMath math="P'(5,0)" />.

2. Given and the mirror line . So .

 <MathContainer>
 <BlockMath math="x' = x = -1" />
 <BlockMath math="y' = 2h - y = 2(0) - (-3) = 0 + 3 = 3" />
 </MathContainer>

 Thus, the image of point is <InlineMath math="Q'(-1,3)" />.

3. Given the image <InlineMath math="R'(-2,5)" /> and the mirror line . So <InlineMath math="x'=-2, y'=5, h=4" />.

 We know <InlineMath math="x' = x" /> and <InlineMath math="y' = 2h - y" />.

 From <InlineMath math="x' = x" />, then .

 From <InlineMath math="y' = 2h - y" />, then .

 <MathContainer>
 
 

 
 

 
 

 </MathContainer>

 Thus, the coordinates of point are .