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URL: https://nakafa.com/en/subjects/mathematics/geometric-transformation/reflection-matrix-arbitrary-point
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Learn how to reflect points over an arbitrary point using translation, matrix operations, and combined transformations.

---

## Finding the Reflection Matrix over an Arbitrary Point

The image of a point $$(x,y)$$ reflected over the point $$(a,b)$$ is $$(-x+2a, -y+2b)$$ or $$(2a-x, 2b-y)$$.

Visible text: The image of a point reflected over the point is or .

The matrix operation associated with this transformation cannot be represented solely by a single $$2 \times 2$$ multiplication matrix, as it involves addition (translation) due to the center point $$(a,b)$$ not being the origin.

Visible text: The matrix operation associated with this transformation cannot be represented solely by a single multiplication matrix, as it involves addition (translation) due to the center point not being the origin.

However, we can represent this transformation as a combination of matrix operations:

1. Translate the point $$(x,y)$$ so that the center of reflection $$(a,b)$$ effectively becomes the origin. This means we work with $$(x-a, y-b)$$.
2. Reflect this translated point over the origin using the matrix $$\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$$.
3. Translate the result back by adding the coordinates of the center of reflection $$(a,b)$$.

Visible text: 1. Translate the point so that the center of reflection effectively becomes the origin. This means we work with .
2. Reflect this translated point over the origin using the matrix .
3. Translate the result back by adding the coordinates of the center of reflection .

Mathematically, if <InlineMath math="(x', y')" /> is the image of $$(x,y)$$:

Visible text: Mathematically, if <InlineMath math="(x', y')" /> is the image of :

Component: MathContainer
Children:

```math
\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} x-a \\ y-b \end{pmatrix} + \begin{pmatrix} a \\ b \end{pmatrix}
```

```math
\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -(x-a) \\ -(y-b) \end{pmatrix} + \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} -x+a \\ -y+b \end{pmatrix} + \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} -x+2a \\ -y+2b \end{pmatrix}
```

This corresponds to the formula we are familiar with.

### Matrix Operation for Reflection over a Point

The matrix operation associated with reflection over the point $$P(a,b)$$ for any point $$(x,y)$$ is:

Visible text: The matrix operation associated with reflection over the point for any point is:

```math
\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} + 2 \begin{pmatrix} a \\ b \end{pmatrix}
```

Or, more precisely, it can be written as a combination:

```math
\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \left( \begin{pmatrix} x \\ y \end{pmatrix} - \begin{pmatrix} a \\ b \end{pmatrix} \right) + \begin{pmatrix} a \\ b \end{pmatrix}
```

The form presented as Property $$4.11$$ in the book ($$\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} + 2 \begin{pmatrix} a \\ b \end{pmatrix}$$) is a simplification of $$\begin{pmatrix} -x+2a \\ -y+2b \end{pmatrix}$$.

Visible text: The form presented as Property in the book () is a simplification of .

## Finding the Image of a Point

Determine the image of point $$(2,3)$$ by reflection over point $$(1,1)$$.

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

**Alternative Solution:**

Point $$(x,y) = (2,3)$$. Center $$(a,b) = (1,1)$$.

Visible text: Point . Center .

Component: MathContainer
Children:

```math
\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix} + 2 \begin{pmatrix} 1 \\ 1 \end{pmatrix}
```

```math
= \begin{pmatrix} (-1)(2) + (0)(3) \\ (0)(2) + (-1)(3) \end{pmatrix} + \begin{pmatrix} 2 \\ 2 \end{pmatrix}
```

```math
= \begin{pmatrix} -2 \\ -3 \end{pmatrix} + \begin{pmatrix} 2 \\ 2 \end{pmatrix} = \begin{pmatrix} -2+2 \\ -3+2 \end{pmatrix} = \begin{pmatrix} 0 \\ -1 \end{pmatrix}
```

Visualization:

Component: LineEquation
Props:
- title: Reflection of Point $$Q(2,3)$$ over{" "}
$$P(1,1)$$
  Visible text: Reflection of Point over{" "}
- description: Point $$Q(2,3)$$ reflected over{" "}
$$P(1,1)$$ becomes <InlineMath math="Q'(0,-1)" />.
  Visible text: Point reflected over{" "}
 becomes <InlineMath math="Q'(0,-1)" />.
- data: [
{
points: [{ x: 1, y: 1, z: 0 }],
color: getColor("ROSE"),
showPoints: true,
labels: [{ text: "P(1,1) - Center", at: 0, offset: [0.3, -0.5, 0] }],
},
{
points: [{ x: 2, y: 3, z: 0 }],
color: getColor("CYAN"),
showPoints: true,
labels: [{ text: "Q(2,3) - Original", at: 0, offset: [0.3, 0.3, 0] }],
},
{
points: [{ x: 0, y: -1, z: 0 }],
color: getColor("EMERALD"),
showPoints: true,
labels: [{ text: "Q'(0,-1) - Image", at: 0, offset: [-0.8, -0.2, 0] }],
},
{
points: [
{ x: 2, y: 3, z: 0 },
{ x: 0, y: -1, z: 0 },
],
color: getColor("INDIGO"),
}, // Line QQ'
]
- showZAxis: false
- cameraPosition: [1, 1, 10]

## Exercises

1.  Determine the image of point $$(3,1)$$ by reflection over point $$(-2,-1)$$.
2.  A line passes through points $$A(1,2)$$ and $$B(3,4)$$. Determine the equation of the image line after reflection over point $$C(0,1)$$.

Visible text: 1. Determine the image of point by reflection over point .
2. A line passes through points and . Determine the equation of the image line after reflection over point .

### Key Answers

1.  Point $$(x,y) = (3,1)$$. Center $$(a,b) = (-2,-1)$$.

    Using the formula <InlineMath math="x' = 2a-x" /> and <InlineMath math="y' = 2b-y" />:

    <MathContainer>
      <BlockMath math="x' = 2(-2) - 3 = -4 - 3 = -7" />
      <BlockMath math="y' = 2(-1) - 1 = -2 - 1 = -3" />
    </MathContainer>

    Its image is $$(-7,-3)$$. Or using matrix operations:

    <BlockMath math="\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 3 \\ 1 \end{pmatrix} + 2 \begin{pmatrix} -2 \\ -1 \end{pmatrix} = \begin{pmatrix} -3 \\ -1 \end{pmatrix} + \begin{pmatrix} -4 \\ -2 \end{pmatrix} = \begin{pmatrix} -7 \\ -3 \end{pmatrix}" />

2.  Center of reflection $$C(0,1)$$. ($$a=0, b=1$$)

    Image of point $$A(1,2)$$:

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

    So <InlineMath math="A'(-1,0)" />.

    Image of point $$B(3,4)$$:

    <MathContainer>
      <BlockMath math="x'_B = 2(0)-3 = -3" />
      <BlockMath math="y'_B = 2(1)-4 = -2" />
    </MathContainer>

    So <InlineMath math="B'(-3,-2)" />.

    The image line passes through <InlineMath math="A'(-1,0)" /> and <InlineMath math="B'(-3,-2)" />.

    Gradient <InlineMath math="m' = \frac{-2 - 0}{-3 - (-1)} = \frac{-2}{-2} = 1" />.

    Equation of the line:

    <MathContainer>
      <BlockMath math="y - y'_A = m'(x - x'_A)" />
      
    
    ```math
    y - 0 = 1(x - (-1))
    ```

      
    
    ```math
    y = x + 1
    ```

    </MathContainer>

    or

    
    
    ```math
    x - y + 1 = 0
    ```

Visible text: 1. Point . Center .

 Using the formula <InlineMath math="x' = 2a-x" /> and <InlineMath math="y' = 2b-y" />:

 <MathContainer>
 <BlockMath math="x' = 2(-2) - 3 = -4 - 3 = -7" />
 <BlockMath math="y' = 2(-1) - 1 = -2 - 1 = -3" />
 </MathContainer>

 Its image is . Or using matrix operations:

 <BlockMath math="\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 3 \\ 1 \end{pmatrix} + 2 \begin{pmatrix} -2 \\ -1 \end{pmatrix} = \begin{pmatrix} -3 \\ -1 \end{pmatrix} + \begin{pmatrix} -4 \\ -2 \end{pmatrix} = \begin{pmatrix} -7 \\ -3 \end{pmatrix}" />

2. Center of reflection . ()

 Image of point :

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

 So <InlineMath math="A'(-1,0)" />.

 Image of point :

 <MathContainer>
 <BlockMath math="x'_B = 2(0)-3 = -3" />
 <BlockMath math="y'_B = 2(1)-4 = -2" />
 </MathContainer>

 So <InlineMath math="B'(-3,-2)" />.

 The image line passes through <InlineMath math="A'(-1,0)" /> and <InlineMath math="B'(-3,-2)" />.

 Gradient <InlineMath math="m' = \frac{-2 - 0}{-3 - (-1)} = \frac{-2}{-2} = 1" />.

 Equation of the line:

 <MathContainer>
 <BlockMath math="y - y'_A = m'(x - x'_A)" />
 
 

 
 

 </MathContainer>

 or