# Nakafa Framework: LLM

URL: /en/subject/high-school/11/mathematics/geometric-transformation/translation
Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/subject/high-school/11/mathematics/geometric-transformation/translation/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",
  description: "Master geometric translation transformations: learn to translate points and lines using vectors with step-by-step examples and visual demonstrations.",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "05/10/2025",
  subject: "Geometric Transformation",
};

## Understanding Translation

Translation, also known as a shift or slide, is a type of geometric transformation that moves every point of an object a certain distance in a specified direction. This transformation does not change the orientation, size, or shape of the object; only its position changes.

### Definition of Translation

Given any point <InlineMath math="P(x,y)" />. The translation associated with the vector <InlineMath math="\begin{pmatrix} a \\ b \end{pmatrix}" /> for point <InlineMath math="P(x,y)" />, written as <InlineMath math="T_{(a,b)}(x,y)" /> or <InlineMath math="\tau_{(a,b)}(x,y)" />, is defined as:

<BlockMath math="P'(x',y') = (x+a, y+b)" />

This means:

<div className="flex flex-col gap-4">
  <BlockMath math="x' = x + a" />
  <BlockMath math="y' = y + b" />
</div>

Here, <InlineMath math="a" /> is the horizontal shift (positive to the right, negative to the left) and <InlineMath math="b" /> is the vertical shift (positive upwards, negative downwards).

## Translating a Point

A point <InlineMath math="(3,2)" /> is translated by the vector <InlineMath math="\begin{pmatrix} -2 \\ 3 \end{pmatrix}" />. Determine the image point of this translation.

Here, <InlineMath math="x=3" />, <InlineMath math="y=2" />, <InlineMath math="a=-2" />, and <InlineMath math="b=3" />.

Using the formula:

<div className="flex flex-col gap-4">
  <BlockMath math="x' = x + a = 3 + (-2) = 1" />
  <BlockMath math="y' = y + b = 2 + 3 = 5" />
</div>

Thus, the image of point <InlineMath math="(3,2)" /> is <InlineMath math="(1,5)" />.

<LineEquation
  title={
    <>
      Translation of Point <InlineMath math="P(3,2)" /> by Vector{" "}
      <InlineMath math="\begin{pmatrix} -2 \\ 3 \end{pmatrix}" />
    </>
  }
  description={
    <>
      Visualization of translating point <InlineMath math="P(3,2)" /> to{" "}
      <InlineMath math="P'(1,5)" /> using a translation vector.
    </>
  }
  data={[
    {
      points: [{ x: 3, y: 2, z: 0 }],
      color: getColor("SKY"),
      showPoints: true,
      labels: [{ text: "P(3,2) - Original", at: 0, offset: [0.3, -0.3, 0] }],
    }, // Original Point
    {
      points: [{ x: 1, y: 5, z: 0 }],
      color: getColor("EMERALD"),
      showPoints: true,
      labels: [{ text: "P'(1,5) - Image", at: 0, offset: [0.3, 0.3, 0] }],
    }, // Image Point
    {
      points: [
        { x: 3, y: 2, z: 0 },
        { x: 1, y: 5, z: 0 },
      ],
      color: getColor("ROSE"),
      labels: [{ text: "vector (-2,3)", at: 0, offset: [-1, 1.5, 0] }],
    }, // Translation Vector from P to P'
  ]}
  showZAxis={false}
/>

## Translating a Line

Determine the image of the line <InlineMath math="l \equiv 2x + 3y - 1 = 0" /> translated by the vector <InlineMath math="\begin{pmatrix} -1 \\ 1 \end{pmatrix}" />.

Let <InlineMath math="P(x,y)" /> be any point on line <InlineMath math="l" />. If translated by the vector <InlineMath math="\begin{pmatrix} -1 \\ 1 \end{pmatrix}" />, its image is <InlineMath math="P'(x',y')" /> where:

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

Substitute these values of <InlineMath math="x" /> and <InlineMath math="y" /> into the equation of line <InlineMath math="l" />:

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

Replacing <InlineMath math="x'" /> and <InlineMath math="y'" /> back to <InlineMath math="x" /> and <InlineMath math="y" />, the equation of the image line <InlineMath math="l'" /> is:

<div className="flex flex-col gap-4">
  <BlockMath math="2x + 3y - 2 = 0" />
  <LineEquation
    title={
      <>
        Translation of Line <InlineMath math="2x+3y-1=0" /> by Vector{" "}
        <InlineMath math="\begin{pmatrix} -1 \\ 1 \end{pmatrix}" />
      </>
    }
    description={
      <>
        Original line <InlineMath math="2x+3y-1=0" /> translated results in
        image line <InlineMath math="2x+3y-2=0" />.
      </>
    }
    data={[
      {
        // Original Line: 2x + 3y - 1 = 0  => y = (-2/3)x + 1/3
        points: Array.from({ length: 11 }, (_, i) => {
          const xVal = i - 5;
          return { x: xVal, y: (-2 / 3) * xVal + 1 / 3, z: 0 };
        }),
        color: getColor("PURPLE"),
        labels: [{ text: "2x+3y-1=0", at: 2, offset: [-1, -0.5, 0] }],
      },
      {
        // Image Line: 2x + 3y - 2 = 0 => y = (-2/3)x + 2/3
        points: Array.from({ length: 11 }, (_, i) => {
          const xVal = i - 5;
          return { x: xVal, y: (-2 / 3) * xVal + 2 / 3, z: 0 };
        }),
        color: getColor("PINK"),
        labels: [{ text: "2x+3y-2=0", at: 8, offset: [1, 0.5, 0] }],
      },
    ]}
    showZAxis={false}
    cameraPosition={[0, 0, 10]}
  />
</div>

## Exercises

1.  A point <InlineMath math="(-4,4)" /> is translated by the vector <InlineMath math="\begin{pmatrix} -2 \\ -3 \end{pmatrix}" />. Determine the image point of this translation.
2.  Determine the image of the line <InlineMath math="l' \equiv 5x - 2y + 3 = 0" /> translated by the vector <InlineMath math="\begin{pmatrix} 2 \\ -1 \end{pmatrix}" />.
3.  A triangle with vertices <InlineMath math="A(1,1)" />, <InlineMath math="B(4,1)" />, and <InlineMath math="C(1,5)" /> is translated by the vector <InlineMath math="\begin{pmatrix} 2 \\ 3 \end{pmatrix}" />. Determine the coordinates of the image triangle <InlineMath math="A'B'C'" />!

### Key Answers

1.  Point <InlineMath math="(-4,4)" />, vector <InlineMath math="\begin{pmatrix} -2 \\ -3 \end{pmatrix}" />. <InlineMath math="x=-4, y=4, a=-2, b=-3" />.

    <div className="flex flex-col gap-4">
      <BlockMath math="x' = -4 + (-2) = -6" />
      <BlockMath math="y' = 4 + (-3) = 1" />
    </div>

    Thus, the image point is <InlineMath math="(-6,1)" />.

2.  Line <InlineMath math="5x - 2y + 3 = 0" />, vector <InlineMath math="\begin{pmatrix} 2 \\ -1 \end{pmatrix}" />. <InlineMath math="a=2, b=-1" />.

    <div className="flex flex-col gap-4">
      <BlockMath math="x' = x + 2 \implies x = x' - 2" />
      <BlockMath math="y' = y - 1 \implies y = y' + 1" />
    </div>

    Substitute into the line equation:

    <div className="flex flex-col gap-4">
      <BlockMath math="5(x' - 2) - 2(y' + 1) + 3 = 0" />
      <BlockMath math="5x' - 10 - 2y' - 2 + 3 = 0" />
      <BlockMath math="5x' - 2y' - 9 = 0" />
    </div>

    Image line equation: <InlineMath math="5x - 2y - 9 = 0" />.

3.  Points <InlineMath math="A(1,1)" />, <InlineMath math="B(4,1)" />, <InlineMath math="C(1,5)" />. Vector <InlineMath math="\begin{pmatrix} 2 \\ 3 \end{pmatrix}" />.

    <div className="flex flex-col gap-4">
      <BlockMath math="A'(1+2, 1+3) = A'(3,4)" />
      <BlockMath math="B'(4+2, 1+3) = B'(6,4)" />
      <BlockMath math="C'(1+2, 5+3) = C'(3,8)" />
    </div>

    The coordinates of the image triangle are <InlineMath math="A'(3,4)" />, <InlineMath math="B'(6,4)" />, and <InlineMath math="C'(3,8)" />.
