Translation Matrix

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a

b

(x+a, y+b)

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

\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}

Rotations about the origin and reflections in lines through the origin can be represented by $2 \times 2$ matrix multiplication. A nonzero translation instead adds a vector and $cannot be represented this way$ .

Why Homogeneous Coordinates Are Needed

Ordinary matrix multiplication always sends the $origin to the origin$ , while a nonzero translation moves it. This is why we need an extra coordinate:

M\begin{pmatrix}0\\0\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix},\qquad T_{(a,b)}(0,0)=(a,b)

To combine translation with linear transformations using matrix multiplication, we use $homogeneous coordinates$ . The point $(x,y)$ is represented by the column:

\begin{pmatrix} x \\ y \\ 1 \end{pmatrix}

The transformation then uses a $3 \times 3$ matrix. For a horizontal shift of $a$ and a vertical shift of $b$ , it is:

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

The calculation gives:

\begin{aligned} \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} \end{aligned}

Checking Homogeneous Coordinates

The last coordinate stays $equal to one$ . In the example below, the first row calculates horizontal position and the second row calculates vertical position. The result agrees with vector addition:

\begin{pmatrix}1&0&-3\\0&1&4\\0&0&1\end{pmatrix}\begin{pmatrix}-2\\3\\1\end{pmatrix}=\begin{pmatrix}1(-2)+0(3)-3(1)\\0(-2)+1(3)+4(1)\\1\end{pmatrix}=\begin{pmatrix}-5\\7\\1\end{pmatrix}

Finding the Image of a Point with Matrix Operation

Use matrix operations to translate the point $(-2,3)$ by the following vector and determine its image:

\begin{pmatrix} -3 \\ 4 \end{pmatrix}

$Solution$ :

Add the $corresponding components$ directly:

\begin{aligned} \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} \end{aligned}

Its image is $(-5,7)$ .

Translation from

P(-2,3)

to

P'(-5,7)

Visualization of translating point

P(-2,3)

to

P'(-5,7)

by a translation vector.

Exercises

Apply the $same vector$ to every point. Put the horizontal component in the first row and the vertical component in the second. For the triangle, repeat the addition for all three vertices.

Use matrix operations to translate the point $(4,-5)$ by the following vector and determine its image:

$\begin{pmatrix} -5 \\ 4 \end{pmatrix}$
A triangle $KLM$ has vertices $K(1,0)$ , $L(4,2)$ , and $M(2,5)$ . Translate it by the vector below and determine the coordinates of $K'L'M'$ :

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

Solutions

Add the $corresponding components$ . After finding each image, subtract its original coordinates to check that the resulting displacement equals the vector given in the problem.

$Solution 1$

For the point $(4,-5)$ , use the shifts $a=-5$ and $b=4$ .

\begin{aligned} \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} \end{aligned}

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

$Solution 2$

Apply the given translation vector $T$ to each vertex:

For $K(1,0)$ :

\begin{aligned} \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} \end{aligned}

So $K'(-1,1)$ .

For $L(4,2)$ :

\begin{aligned} \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} \end{aligned}

So $L'(2,3)$ .

For $M(2,5)$ :

\begin{aligned} \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} \end{aligned}

So $M'(0,6)$ .

Image coordinates: $K'(-1,1)$ , $L'(2,3)$ , $M'(0,6)$ .

Translation Matrix

Moving Points by Adding a Vector

Translation Matrix

Moving Points by Adding a Vector

Why Homogeneous Coordinates Are Needed

Checking Homogeneous Coordinates

Finding the Image of a Point with Matrix Operation

Exercises

Solutions