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URL: https://nakafa.com/en/subjects/mathematics/geometric-transformation/reflection-over-x-equals-k
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Learn reflection over vertical lines with worked examples, point-image rules, and practice problems.

---

## Understanding Reflection over a Vertical Line

Reflection over the vertical line $$x = k$$ is a geometric transformation where each point of an object is mapped to a new position. The line $$x = k$$ acts as a mirror.

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

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

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

### Rule for Reflection over a Vertical Line

If a point $$P(x, y)$$ is reflected over the line $$x = k$$, 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' = 2k - x
```

```math
y' = y
```

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

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

## Reflecting a Point over a Vertical Line

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

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

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

Visible text: In this case, , , and .

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

Component: MathContainer
Children:

```math
x' = 2(3) - 3 = 6 - 3 = 3
```

```math
y' = 2
```

Thus, the image of point $$P(3,2)$$ is <InlineMath math="P'(3,2)" />. The point lies on the mirror line, so its image is the point itself.

Visible text: Thus, the image of point is <InlineMath math="P'(3,2)" />. The point lies on the mirror line, so its image is the point itself.

Now, let's try another example. Determine the image of point $$A(1,4)$$ if it is reflected over the line $$x=3$$.

Visible text: Now, let's try another example. Determine the image of point if it is reflected over the line .

Here, $$x=1$$, $$y=4$$, and $$k=3$$.

Visible text: Here, , , and .

Component: MathContainer
Children:

```math
x' = 2(3) - 1 = 6 - 1 = 5
```

```math
y' = 4
```

Thus, the image of point $$A(1,4)$$ is <InlineMath math="A'(5,4)" />.

Visible text: Thus, the image of point is <InlineMath math="A'(5,4)" />.

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

## Exercises

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

Visible text: 1. Determine the image of point by reflection over the line .
2. A point is reflected over the line . Determine the coordinates of its image!
3. The image of a point after reflection over the line is <InlineMath math="C'(0,3)" />. Determine the coordinates of point !

### Key Answers

1.  Given $$P(3,2)$$ and the mirror line $$x=5$$. So $$x=3, y=2, k=5$$.

    <MathContainer>
      <BlockMath math="x' = 2k - x = 2(5) - 3 = 10 - 3 = 7" />
      <BlockMath math="y' = y = 2" />
    </MathContainer>

    Thus, the image of point $$P$$ is <InlineMath math="P'(7,2)" />.

2.  Given $$B(-2, 5)$$ and the mirror line $$x=1$$. So $$x=-2, y=5, k=1$$.

    <MathContainer>
      <BlockMath math="x' = 2k - x = 2(1) - (-2) = 2 + 2 = 4" />
      <BlockMath math="y' = y = 5" />
    </MathContainer>

    Thus, the image of point $$B$$ is <InlineMath math="B'(4,5)" />.

3.  Given the image <InlineMath math="C'(0,3)" /> and the mirror line $$x=-2$$. So <InlineMath math="x'=0, y'=3, k=-2" />.

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

    From <InlineMath math="y' = y" />, then $$y = 3$$.

    From <InlineMath math="x' = 2k - x" />, then $$0 = 2(-2) - x$$.

    <MathContainer>
      
    
    ```math
    0 = -4 - x
    ```

      
    
    ```math
    x = -4
    ```

    </MathContainer>

    Thus, the coordinates of point $$C$$ are $$C(-4,3)$$.

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

 <MathContainer>
 <BlockMath math="x' = 2k - x = 2(5) - 3 = 10 - 3 = 7" />
 <BlockMath math="y' = y = 2" />
 </MathContainer>

 Thus, the image of point is <InlineMath math="P'(7,2)" />.

2. Given and the mirror line . So .

 <MathContainer>
 <BlockMath math="x' = 2k - x = 2(1) - (-2) = 2 + 2 = 4" />
 <BlockMath math="y' = y = 5" />
 </MathContainer>

 Thus, the image of point is <InlineMath math="B'(4,5)" />.

3. Given the image <InlineMath math="C'(0,3)" /> and the mirror line . So <InlineMath math="x'=0, y'=3, k=-2" />.

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

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

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

 <MathContainer>
 
 

 
 

 </MathContainer>

 Thus, the coordinates of point are .