# Nakafa Framework: LLM

URL: /en/subject/high-school/11/mathematics/geometric-transformation/reflection-matrix-center
Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/subject/high-school/11/mathematics/geometric-transformation/reflection-matrix-center/en.mdx

Output docs content for large language models.

---

import { getColor } from "@repo/design-system/lib/color";
import { LineEquation } from "@repo/design-system/components/contents/line-equation";

export const metadata = {
  title: "Reflection Matrix over Center Point",
  description: "Learn reflection matrix over origin (0,0). Master point transformation using 2x2 matrices with step-by-step calculations and visual examples.",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "05/10/2025",
  subject: "Geometric Transformation",
};

## Finding the Reflection Matrix over the Origin

Reflecting a point <InlineMath math="(x,y)" /> over the origin <InlineMath math="O(0,0)" /> results in the image <InlineMath math="(-x,-y)" />. This is equivalent to a <InlineMath math="180^\circ" /> rotation about the origin.

Now, we will find the <InlineMath math="2 \times 2" /> matrix, let's say <InlineMath math="\begin{pmatrix} r & s \\ t & u \end{pmatrix}" />, that represents this transformation.

We want to find <InlineMath math="r, s, t, u" /> such that:

<BlockMath 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:

<BlockMath 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: <InlineMath math="rx + sy = -1x + 0y" />. This means <InlineMath math="r = -1" /> and <InlineMath math="s = 0" />.
- For the second row: <InlineMath math="tx + uy = 0x - 1y" />. This means <InlineMath math="t = 0" /> and <InlineMath math="u = -1" />.

### Reflection Matrix over the Origin

The matrix associated with reflection over the origin <InlineMath math="O(0,0)" /> is:

<BlockMath 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 <InlineMath math="A(-1,1)" /> and <InlineMath math="B(3,-2)" /> when reflected over the origin!

**Alternative Solution:**

Using the transformation matrix <InlineMath math="\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}" />:

For point <InlineMath math="A(-1,1)" />:

<BlockMath 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)" />.

For point <InlineMath math="B(3,-2)" />:

<BlockMath 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)" />.

<LineEquation
  title={
    <>
      Reflection of Points <InlineMath math="A" /> and <InlineMath math="B" />{" "}
      over the Origin
    </>
  }
  description={
    <>
      Visualization of reflecting point <InlineMath math="A(-1,1)" /> to{" "}
      <InlineMath math="A'(1,-1)" /> and <InlineMath math="B(3,-2)" /> to{" "}
      <InlineMath math="B'(-3,2)" /> over the origin{" "}
      <InlineMath math="O(0,0)" />.
    </>
  }
  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 <InlineMath math="A(1,-7)" /> and <InlineMath math="B(-7,-2)" /> when reflected over the origin!
2.  A triangle <InlineMath math="PQR" /> has vertices <InlineMath math="P(2,2)" />, <InlineMath math="Q(5,2)" />, and <InlineMath math="R(3,5)" />. 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: <InlineMath math="\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}" />.

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

    <BlockMath 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 <InlineMath math="B(-7,-2)" />:

    <BlockMath 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: <InlineMath math="\begin{pmatrix} 2 & 5 & 3 \\ 2 & 2 & 5 \end{pmatrix}" />.

    <BlockMath 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)" />.
