Translation Matrix

Matrix Operation for Translation

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

Translation Matrix

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)

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}

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.

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:

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

Thus:

\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:

\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.

Alternative Solution:

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

\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 $(-5,7)$ .

Translation of Point

P(-2,3)

by Vector

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

Visualization of translating point

P(-2,3)

to

P'(-5,7)

by a translation vector.

Exercises

Determine the image of point $(4,-5)$ translated by the vector $\begin{pmatrix} -5 \\ 4 \end{pmatrix}$ using matrix operation.
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 $K'L'M'$ .

Key Answers

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

$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 4 \\ -5 \end{pmatrix} + \begin{pmatrix} -5 \\ 4 \end{pmatrix} = \begin{pmatrix} 4-5 \\ -5+4 \end{pmatrix} = \begin{pmatrix} -1 \\ -1 \end{pmatrix}$

Image: $(-1,-1)$ .
Translation vector $T = \begin{pmatrix} -2 \\ 1 \end{pmatrix}$ .
- For $K(1,0)$ : $\begin{pmatrix} x'_K \\ y'_K \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \begin{pmatrix} -2 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}$ . So .

K'(-1,1)

For

L(4,2)

:

\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

L'(2,3)

.

For

M(2,5)

:

\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

M'(0,6)

.

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