# Nakafa Learning Content

> For AI agents: use [llms.txt](https://nakafa.com/llms.txt) for the site index. Markdown versions are available by appending `.md` to content URLs or sending `Accept: text/markdown`.

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

Learn reflection matrices over the origin. Learn point transformations with worked calculations and visual examples.

---

## Finding the Reflection Matrix over the Origin

Reflecting a point $$(x,y)$$ over the origin $$O(0,0)$$ results in the image $$(-x,-y)$$. This is equivalent to a $$180^\circ$$ rotation about the origin.

Visible text: Reflecting a point over the origin results in the image . This is equivalent to a rotation about the origin.

Now, we will find the $$2 \times 2$$ matrix, let's say $$\begin{pmatrix} r & s \\ t & u \end{pmatrix}$$, that represents this transformation.

Visible text: Now, we will find the matrix, let's say , that represents this transformation.

We want to find $$r, s, t, u$$ such that:

Visible text: We want to find such that:

```math
\begin{pmatrix} r & s \\ t & u \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -x \\ -y \end{pmatrix}
```

From matrix multiplication, we can write:

```math
\begin{pmatrix} rx + sy \\ tx + uy \end{pmatrix} = \begin{pmatrix} -1x + 0y \\ 0x - 1y \end{pmatrix}
```

By equating the corresponding coefficients, we get:

- For the first row: $$rx + sy = -1x + 0y$$. This means $$r = -1$$ and $$s = 0$$.
- For the second row: $$tx + uy = 0x - 1y$$. This means $$t = 0$$ and $$u = -1$$.

Visible text: - For the first row: . This means and .
- For the second row: . This means and .

### Reflection Matrix over the Origin

The matrix associated with reflection over the origin $$O(0,0)$$ is:

Visible text: The matrix associated with reflection over the origin is:

```math
\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}
```

## Application of Reflection Matrix over the Origin

## Finding the Image of Points

Determine the images of points $$A(-1,1)$$ and $$B(3,-2)$$ when reflected over the origin!

Visible text: Determine the images of points and when reflected over the origin!

**Alternative Solution:**

Using the transformation matrix $$\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$$:

Visible text: Using the transformation matrix :

For point $$A(-1,1)$$:

Visible text: For point :

```math
\begin{pmatrix} x'_A \\ y'_A \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{pmatrix} (-1)(-1) + (0)(1) \\ (0)(-1) + (-1)(1) \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix}
```

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

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

For point $$B(3,-2)$$:

Visible text: For point :

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

The image of point $$B$$ is <InlineMath math="B'(-3,2)" />.

Visible text: The image of point is <InlineMath math="B'(-3,2)" />.

Component: LineEquation
Props:
- title: Reflection of Points $$A$$ and $$B$${" "}
over the Origin
  Visible text: Reflection of Points and {" "}
over the Origin
- description: Visualization of reflecting point $$A(-1,1)$$ to{" "}
<InlineMath math="A'(1,-1)" /> and $$B(3,-2)$$ to{" "}
<InlineMath math="B'(-3,2)" /> over the origin{" "}
$$O(0,0)$$.
  Visible text: Visualization of reflecting point to{" "}
<InlineMath math="A'(1,-1)" /> and to{" "}
<InlineMath math="B'(-3,2)" /> over the origin{" "}
.
- data: [
{
points: [{ x: 0, y: 0, z: 0 }],
color: getColor("ROSE"),
showPoints: true,
labels: [{ text: "O", at: 0, offset: [0.3, -0.3, 0] }],
}, // Origin
// Point A and A'
{
points: [{ x: -1, y: 1, z: 0 }],
color: getColor("CYAN"),
showPoints: true,
labels: [{ text: "A(-1,1)", at: 0, offset: [-0.7, 0.3, 0] }],
},
{
points: [{ x: 1, y: -1, z: 0 }],
color: getColor("CYAN"),
showPoints: true,
labels: [{ text: "A'(1,-1)", at: 0, offset: [0.3, -0.5, 0] }],
},
{
points: [
{ x: -1, y: 1, z: 0 },
{ x: 1, y: -1, z: 0 },
],
color: getColor("INDIGO"),
}, // Line AA'
// Point B and B'
{
points: [{ x: 3, y: -2, z: 0 }],
color: getColor("PURPLE"),
showPoints: true,
labels: [{ text: "B(3,-2)", at: 0, offset: [0.3, -0.3, 0] }],
},
{
points: [{ x: -3, y: 2, z: 0 }],
color: getColor("PURPLE"),
showPoints: true,
labels: [{ text: "B'(-3,2)", at: 0, offset: [-0.7, 0.3, 0] }],
},
{
points: [
{ x: 3, y: -2, z: 0 },
{ x: -3, y: 2, z: 0 },
],
color: getColor("INDIGO"),
}, // Line BB'
]
- showZAxis: false
- cameraPosition: [0, 0, 10]

## Exercises

1.  Determine the images of points $$A(1,-7)$$ and $$B(-7,-2)$$ when reflected over the origin!
2.  A triangle $$PQR$$ has vertices $$P(2,2)$$, $$Q(5,2)$$, and $$R(3,5)$$. Determine the coordinates of the image triangle <InlineMath math="P'Q'R'" /> after reflection over the origin using matrix multiplication.

Visible text: 1. Determine the images of points and when reflected over the origin!
2. A triangle has vertices , , and . Determine the coordinates of the image triangle <InlineMath math="P'Q'R'" /> after reflection over the origin using matrix multiplication.

### Key Answers

1.  The reflection matrix over the origin is: $$\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$$.

    For $$A(1,-7)$$:

    
    
    ```math
    \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 1 \\ -7 \end{pmatrix} = \begin{pmatrix} -1 \\ 7 \end{pmatrix}
    ```

    Image <InlineMath math="A'(-1,7)" />. For $$B(-7,-2)$$:

    
    
    ```math
    \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} -7 \\ -2 \end{pmatrix} = \begin{pmatrix} 7 \\ 2 \end{pmatrix}
    ```

    Image <InlineMath math="B'(7,2)" />.

2.  Matrix of PQR vertices: $$\begin{pmatrix} 2 & 5 & 3 \\ 2 & 2 & 5 \end{pmatrix}$$.

    
    
    ```math
    \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 2 & 5 & 3 \\ 2 & 2 & 5 \end{pmatrix} = \begin{pmatrix} -2 & -5 & -3 \\ -2 & -2 & -5 \end{pmatrix}
    ```

    Image: <InlineMath math="P'(-2,-2)" />, <InlineMath math="Q'(-5,-2)" />, <InlineMath math="R'(-3,-5)" />.

Visible text: 1. The reflection matrix over the origin is: .

 For :

 
 

 Image <InlineMath math="A'(-1,7)" />. For :

 
 

 Image <InlineMath math="B'(7,2)" />.

2. Matrix of PQR vertices: .

 
 

 Image: <InlineMath math="P'(-2,-2)" />, <InlineMath math="Q'(-5,-2)" />, <InlineMath math="R'(-3,-5)" />.