Translation Matrix

Moving Points by Adding a Vector

A translation moves a point $(x,y)$ horizontally by $a$ and vertically by $b$, giving the image $(x+a, y+b)$.

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

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.

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

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

The calculation gives:

Matrix Operation

For a point $(x,y)$, add the translation vector component by component:

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:

Solution:

Apply the vector addition directly:

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

Exercises

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

   $$\begin{pmatrix} -5 \\ 4 \end{pmatrix}$$

2. 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

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: $(-1,-1)$.

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