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):

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:

Thus:

Matrix Operation

The matrix operation associated with translation by vector $\begin{pmatrix} a \\ b \end{pmatrix}$ for point $(x,y)$ is:

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:

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

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 $K'L'M'$.

Key Answers

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

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