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

Learn geometric translation transformations by moving points and lines with vectors, worked examples, and visual demonstrations.

---

## 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 $$P(x,y)$$. The translation associated with the vector $$\begin{pmatrix} a \\ b \end{pmatrix}$$ for point $$P(x,y)$$, written as $$T_{(a,b)}(x,y)$$ or $$\tau_{(a,b)}(x,y)$$, is defined as:

Visible text: Given any point . The translation associated with the vector for point , written as or , is defined as:

```math
P'(x',y') = (x+a, y+b)
```

This means:

Component: MathContainer
Children:

```math
x' = x + a
```

```math
y' = y + b
```

Here, $$a$$ is the horizontal shift (positive to the right, negative to the left) and $$b$$ is the vertical shift (positive upwards, negative downwards).

Visible text: Here, is the horizontal shift (positive to the right, negative to the left) and is the vertical shift (positive upwards, negative downwards).

## Translating a Point

A point $$(3,2)$$ is translated by the vector $$\begin{pmatrix} -2 \\ 3 \end{pmatrix}$$. Determine the image point of this translation.

Visible text: A point is translated by the vector . Determine the image point of this translation.

Here, $$x=3$$, $$y=2$$, $$a=-2$$, and $$b=3$$.

Visible text: Here, , , , and .

Using the formula:

Component: MathContainer
Children:

```math
x' = x + a = 3 + (-2) = 1
```

```math
y' = y + b = 2 + 3 = 5
```

Thus, the image of point $$(3,2)$$ is $$(1,5)$$.

Visible text: Thus, the image of point is .

Component: LineEquation
Props:
- title: Translation of Point $$P(3,2)$$ by Vector{" "}
$$\begin{pmatrix} -2 \\ 3 \end{pmatrix}$$
  Visible text: Translation of Point by Vector{" "}
- description: Visualization of translating point $$P(3,2)$$ to{" "}
<InlineMath math="P'(1,5)" /> using a translation vector.
  Visible text: Visualization of translating point 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 $$l \equiv 2x + 3y - 1 = 0$$ translated by the vector $$\begin{pmatrix} -1 \\ 1 \end{pmatrix}$$.

Visible text: Determine the image of the line translated by the vector .

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

Visible text: Let be any point on line . If translated by the vector , its image is <InlineMath math="P'(x',y')" /> where:

Component: MathContainer
Children:

```math
x' = x + (-1) = x - 1 \implies x = x' + 1
```

```math
y' = y + 1 \implies y = y' - 1
```

Substitute these values of $$x$$ and $$y$$ into the equation of line $$l$$:

Visible text: Substitute these values of and into the equation of line :

Component: MathContainer
Children:

```math
2(x' + 1) + 3(y' - 1) - 1 = 0
```

```math
2x' + 2 + 3y' - 3 - 1 = 0
```

```math
2x' + 3y' - 2 = 0
```

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

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

Component: ContentStack
Children:

```math
2x + 3y - 2 = 0
```

Component: LineEquation
Props:
- title: Translation of Line $$2x+3y-1=0$$ by Vector{" "}
$$\begin{pmatrix} -1 \\ 1 \end{pmatrix}$$
  Visible text: Translation of Line by Vector{" "}
- description: Original line $$2x+3y-1=0$$ translated results in
image line $$2x+3y-2=0$$.
  Visible text: Original line translated results in
image line .
- 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]

## Exercises

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

Visible text: 1. A point is translated by the vector . 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 .
3. A triangle with vertices , , and is translated by the vector . Determine the coordinates of the image triangle <InlineMath math="A'B'C'" />!

### Key Answers

1.  Point $$(-4,4)$$, vector $$\begin{pmatrix} -2 \\ -3 \end{pmatrix}$$. $$x=-4, y=4, a=-2, b=-3$$.

    <MathContainer>
      <BlockMath math="x' = -4 + (-2) = -6" />
      <BlockMath math="y' = 4 + (-3) = 1" />
    </MathContainer>

    Thus, the image point is $$(-6,1)$$.

2.  Line $$5x - 2y + 3 = 0$$, vector $$\begin{pmatrix} 2 \\ -1 \end{pmatrix}$$. $$a=2, b=-1$$.

    <MathContainer>
      <BlockMath math="x' = x + 2 \implies x = x' - 2" />
      <BlockMath math="y' = y - 1 \implies y = y' + 1" />
    </MathContainer>

    Substitute into the line equation:

    <MathContainer>
      <BlockMath math="5(x' - 2) - 2(y' + 1) + 3 = 0" />
      <BlockMath math="5x' - 10 - 2y' - 2 + 3 = 0" />
      <BlockMath math="5x' - 2y' - 9 = 0" />
    </MathContainer>

    Image line equation: $$5x - 2y - 9 = 0$$.

3.  Points $$A(1,1)$$, $$B(4,1)$$, $$C(1,5)$$. Vector $$\begin{pmatrix} 2 \\ 3 \end{pmatrix}$$.

    <MathContainer>
      <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)" />
    </MathContainer>

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

Visible text: 1. Point , vector . .

 <MathContainer>
 <BlockMath math="x' = -4 + (-2) = -6" />
 <BlockMath math="y' = 4 + (-3) = 1" />
 </MathContainer>

 Thus, the image point is .

2. Line , vector . .

 <MathContainer>
 <BlockMath math="x' = x + 2 \implies x = x' - 2" />
 <BlockMath math="y' = y - 1 \implies y = y' + 1" />
 </MathContainer>

 Substitute into the line equation:

 <MathContainer>
 <BlockMath math="5(x' - 2) - 2(y' + 1) + 3 = 0" />
 <BlockMath math="5x' - 10 - 2y' - 2 + 3 = 0" />
 <BlockMath math="5x' - 2y' - 9 = 0" />
 </MathContainer>

 Image line equation: .

3. Points , , . Vector .

 <MathContainer>
 <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)" />
 </MathContainer>

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