# Nakafa Framework: LLM

URL: https://nakafa.com/en/subject/high-school/11/mathematics/geometric-transformation/reflection-matrix-arbitrary-point
Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/subject/high-school/11/mathematics/geometric-transformation/reflection-matrix-arbitrary-point/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 Arbitrary Point",
  description: "Discover how to reflect points over any arbitrary point using translation and matrix operations. Master combined transformations with examples.",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "05/10/2025",
  subject: "Geometric Transformation",
};

## Finding the Reflection Matrix over an Arbitrary Point

The image of a point <InlineMath math="(x,y)" /> reflected over the point <InlineMath math="(a,b)" /> is <InlineMath math="(-x+2a, -y+2b)" /> or <InlineMath math="(2a-x, 2b-y)" />.

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

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

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

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

<div className="flex flex-col gap-4">
  <BlockMath 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}" />
  <BlockMath 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}" />
</div>

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 <InlineMath math="P(a,b)" /> for any point <InlineMath math="(x,y)" /> is:

<BlockMath 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:

<BlockMath 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 (<InlineMath math="\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 <InlineMath math="\begin{pmatrix} -x+2a \\ -y+2b \end{pmatrix}" />.

## Finding the Image of a Point

Determine the image of point <InlineMath math="(2,3)" /> by reflection over point <InlineMath math="(1,1)" />.

**Alternative Solution:**

Point <InlineMath math="(x,y) = (2,3)" />. Center <InlineMath math="(a,b) = (1,1)" />.

<div className="flex flex-col gap-4">
  <BlockMath 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}" />
  <BlockMath math="= \begin{pmatrix} (-1)(2) + (0)(3) \\ (0)(2) + (-1)(3) \end{pmatrix} + \begin{pmatrix} 2 \\ 2 \end{pmatrix}" />
  <BlockMath 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}" />
</div>

Visualization:

<LineEquation
  title={
    <>
      Reflection of Point <InlineMath math="Q(2,3)" /> over{" "}
      <InlineMath math="P(1,1)" />
    </>
  }
  description={
    <>
      Point <InlineMath math="Q(2,3)" /> reflected over{" "}
      <InlineMath math="P(1,1)" /> 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 <InlineMath math="(3,1)" /> by reflection over point <InlineMath math="(-2,-1)" />.
2.  A line passes through points <InlineMath math="A(1,2)" /> and <InlineMath math="B(3,4)" />. Determine the equation of the image line after reflection over point <InlineMath math="C(0,1)" />.

### Key Answers

1.  Point <InlineMath math="(x,y) = (3,1)" />. Center <InlineMath math="(a,b) = (-2,-1)" />.

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

    <div className="flex flex-col gap-4">
      <BlockMath math="x' = 2(-2) - 3 = -4 - 3 = -7" />
      <BlockMath math="y' = 2(-1) - 1 = -2 - 1 = -3" />
    </div>

    Its image is <InlineMath math="(-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 <InlineMath math="C(0,1)" />. (<InlineMath math="a=0, b=1" />)

    Image of point <InlineMath math="A(1,2)" />:

    <div className="flex flex-col gap-4">
      <BlockMath math="x'_A = 2(0)-1 = -1" />
      <BlockMath math="y'_A = 2(1)-2 = 0" />
    </div>

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

    Image of point <InlineMath math="B(3,4)" />:

    <div className="flex flex-col gap-4">
      <BlockMath math="x'_B = 2(0)-3 = -3" />
      <BlockMath math="y'_B = 2(1)-4 = -2" />
    </div>

    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:

    <div className="flex flex-col gap-4">
      <BlockMath math="y - y'_A = m'(x - x'_A)" />
      <BlockMath math="y - 0 = 1(x - (-1))" />
      <BlockMath math="y = x + 1" />
    </div>

    or

    <BlockMath math="x - y + 1 = 0" />