# Nakafa Learning Content

> For AI agents: use [llms.txt](https://nakafa.com/llms.txt) for the site index. Markdown versions are available by appending `.md` to content URLs or sending `Accept: text/markdown`.

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

Learn translation matrix operations and homogeneous coordinates: apply vector addition and 3x3 matrices for geometric transformations with examples.

---

## Matrix Operation for Translation

Translation or shifting a point $$(x,y)$$ by a vector $$\begin{pmatrix} a \\ b \end{pmatrix}$$ results in the image $$(x+a, y+b)$$.

Visible text: Translation or shifting a point by a vector results in the image .

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

```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 $$2 \times 2$$ matrix multiplication. Pure translation is a vector addition operation.

Visible text: This is different from transformations like rotation or reflection across an axis/line, which can be represented by 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 $$(x,y)$$ is represented as $$\begin{pmatrix} x \\ y \\ 1 \end{pmatrix}$$, and the transformation matrix becomes $$3 \times 3$$. For translation by $$\begin{pmatrix} a \\ b \end{pmatrix}$$, the matrix is:

Visible text: However, if we want to combine translation with other linear transformations using matrix multiplication, we often use **homogeneous coordinates**. With homogeneous coordinates, a point is represented as , and the transformation matrix becomes . For translation by , the matrix is:

```math
T_{(a,b)} = \begin{pmatrix} 1 & 0 & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}
```

Thus:

```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 $$\begin{pmatrix} a \\ b \end{pmatrix}$$ for point $$(x,y)$$ is:

Visible text: The matrix operation associated with translation by vector for point is:

```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 $$(-2,3)$$ translated by the vector $$\begin{pmatrix} -3 \\ 4 \end{pmatrix}$$ using matrix operation.

Visible text: Determine the image of point translated by the vector using matrix operation.

**Alternative Solution:**

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

```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 $$(-5,7)$$.

Visible text: Its image is .

Component: LineEquation
Props:
- title: Translation of Point $$P(-2,3)$$ by Vector{" "}
$$\begin{pmatrix} -3 \\ 4 \end{pmatrix}$$
  Visible text: Translation of Point by Vector{" "}
- description: Visualization of translating point $$P(-2,3)$$ to{" "}
<InlineMath math="P'(-5,7)" /> by a translation vector.
  Visible text: Visualization of translating point 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 $$(4,-5)$$ translated by the vector $$\begin{pmatrix} -5 \\ 4 \end{pmatrix}$$ using matrix operation.
2.  A triangle $$KLM$$ has vertices $$K(1,0)$$, $$L(4,2)$$, and $$M(2,5)$$. This triangle is translated by vector $$T = \begin{pmatrix} -2 \\ 1 \end{pmatrix}$$. Determine the coordinates of the image triangle <InlineMath math="K'L'M'" />.

Visible text: 1. Determine the image of point translated by the vector using matrix operation.
2. A triangle has vertices , , and . This triangle is translated by vector . Determine the coordinates of the image triangle <InlineMath math="K'L'M'" />.

### Key Answers

1.  Point $$(4,-5)$$, translation vector $$\begin{pmatrix} -5 \\ 4 \end{pmatrix}$$.

    <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: $$(-1,-1)$$.

2.  Translation vector $$T = \begin{pmatrix} -2 \\ 1 \end{pmatrix}$$.

    - For $$K(1,0)$$: <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 $$L(4,2)$$: <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 $$M(2,5)$$: <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)" />.

Visible text: 1. Point , translation vector .

 <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: .

2. Translation vector .

 - For : <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="\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="\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)" />.