# Nakafa Framework: LLM

URL: https://nakafa.com/en/subject/high-school/11/mathematics/geometric-transformation/translation-matrix
Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/subject/high-school/11/mathematics/geometric-transformation/translation-matrix/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: "Translation Matrix",
  description: "Learn translation matrix operations and homogeneous coordinates: apply vector addition and 3x3 matrices for geometric transformations with examples.",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "05/10/2025",
  subject: "Geometric Transformation",
};

## Matrix Operation for Translation

Translation or shifting a point <InlineMath math="(x,y)" /> by a vector <InlineMath math="\begin{pmatrix} a \\ b \end{pmatrix}" /> results in the image <InlineMath math="(x+a, y+b)" />.

This operation can be written in the form of vector addition (column matrix):

<BlockMath math="\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} x+a \\ y+b \end{pmatrix}" />

This is different from transformations like rotation or reflection across an axis/line, which can be represented by <InlineMath math="2 \times 2" /> matrix multiplication. Pure translation is a vector addition operation.

However, if we want to combine translation with other linear transformations using matrix multiplication, we often use **homogeneous coordinates**. With homogeneous coordinates, a point <InlineMath math="(x,y)" /> is represented as <InlineMath math="\begin{pmatrix} x \\ y \\ 1 \end{pmatrix}" />, and the transformation matrix becomes <InlineMath math="3 \times 3" />. For translation by <InlineMath math="\begin{pmatrix} a \\ b \end{pmatrix}" />, the matrix is:

<BlockMath math="T_{(a,b)} = \begin{pmatrix} 1 & 0 & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}" />

Thus:

<BlockMath math="\begin{pmatrix} x' \\ y' \\ 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} = \begin{pmatrix} x+a \\ y+b \\ 1 \end{pmatrix}" />

### Matrix Operation

The matrix operation associated with translation by vector <InlineMath math="\begin{pmatrix} a \\ b \end{pmatrix}" /> for point <InlineMath math="(x,y)" /> is:

<BlockMath math="\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix} + \begin{pmatrix} a \\ b \end{pmatrix}" />

## Finding the Image of a Point with Matrix Operation

Determine the image of point <InlineMath math="(-2,3)" /> translated by the vector <InlineMath math="\begin{pmatrix} -3 \\ 4 \end{pmatrix}" /> using matrix operation.

**Alternative Solution:**

Based on the matrix operation, the image can be determined by:

<BlockMath math="\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} -2 \\ 3 \end{pmatrix} + \begin{pmatrix} -3 \\ 4 \end{pmatrix} = \begin{pmatrix} -2 + (-3) \\ 3 + 4 \end{pmatrix} = \begin{pmatrix} -5 \\ 7 \end{pmatrix}" />

Its image is <InlineMath math="(-5,7)" />.

<LineEquation
  title={
    <>
      Translation of Point <InlineMath math="P(-2,3)" /> by Vector{" "}
      <InlineMath math="\begin{pmatrix} -3 \\ 4 \end{pmatrix}" />
    </>
  }
  description={
    <>
      Visualization of translating point <InlineMath math="P(-2,3)" /> to{" "}
      <InlineMath math="P'(-5,7)" /> by a translation vector.
    </>
  }
  data={[
    {
      points: [{ x: -2, y: 3, z: 0 }],
      color: getColor("CYAN"),
      showPoints: true,
      labels: [{ text: "P(-2,3)", at: 0, offset: [-0.5, -0.5, 0] }],
    },
    {
      points: [{ x: -5, y: 7, z: 0 }],
      color: getColor("EMERALD"),
      showPoints: true,
      labels: [{ text: "P'(-5,7)", at: 0, offset: [-0.5, 0.5, 0] }],
    },
    {
      points: [
        { x: -2, y: 3, z: 0 },
        { x: -5, y: 7, z: 0 },
      ],
      color: getColor("ROSE"),
      lineWidth: 2,
      cone: { position: "end", size: 0.3 },
      labels: [{ text: "vector (-3,4)", at: 0, offset: [-0.5, 2, 0] }],
    },
  ]}
  showZAxis={false}
  cameraPosition={[-10, 10, 12]}
/>

## Exercises

1.  Determine the image of point <InlineMath math="(4,-5)" /> translated by the vector <InlineMath math="\begin{pmatrix} -5 \\ 4 \end{pmatrix}" /> using matrix operation.
2.  A triangle <InlineMath math="KLM" /> has vertices <InlineMath math="K(1,0)" />, <InlineMath math="L(4,2)" />, and <InlineMath math="M(2,5)" />. This triangle is translated by vector <InlineMath math="T = \begin{pmatrix} -2 \\ 1 \end{pmatrix}" />. Determine the coordinates of the image triangle <InlineMath math="K'L'M'" />.

### Key Answers

1.  Point <InlineMath math="(4,-5)" />, translation vector <InlineMath math="\begin{pmatrix} -5 \\ 4 \end{pmatrix}" />.

    <BlockMath math="\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 4 \\ -5 \end{pmatrix} + \begin{pmatrix} -5 \\ 4 \end{pmatrix} = \begin{pmatrix} 4-5 \\ -5+4 \end{pmatrix} = \begin{pmatrix} -1 \\ -1 \end{pmatrix}" />

    Image: <InlineMath math="(-1,-1)" />.

2.  Translation vector <InlineMath math="T = \begin{pmatrix} -2 \\ 1 \end{pmatrix}" />.

    - For <InlineMath math="K(1,0)" />: <InlineMath math="\begin{pmatrix} x'_K \\ y'_K \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \begin{pmatrix} -2 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}" />. So <InlineMath math="K'(-1,1)" />.
    - For <InlineMath math="L(4,2)" />: <InlineMath math="\begin{pmatrix} x'_L \\ y'_L \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix} + \begin{pmatrix} -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}" />. So <InlineMath math="L'(2,3)" />.
    - For <InlineMath math="M(2,5)" />: <InlineMath math="\begin{pmatrix} x'_M \\ y'_M \end{pmatrix} = \begin{pmatrix} 2 \\ 5 \end{pmatrix} + \begin{pmatrix} -2 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 6 \end{pmatrix}" />. So <InlineMath math="M'(0,6)" />.

    Image coordinates: <InlineMath math="K'(-1,1)" />, <InlineMath math="L'(2,3)" />, <InlineMath math="M'(0,6)" />.