Translations

image/svg+xml A B C A' C' B'

We define a translation as an isometry of the euclidean plane characterized by a vector $\vec{u}$, such that it maps $A$ to every point $A'$ on the plane satisfying that:

$$\begin{array}{rcl} T: E & \rightarrow & E \\ A & \rightarrow & A'=T(A)=A+\vec{u} \end{array}$$

or, what is the same, that its associated system of equations carries out:

$$\begin{pmatrix} x'_1 \\ x'_2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} $$

By the correspondence between points, we can understand the translations as direct movements without changes of orientation, that is to say, they keep the same form and the size of the figures or translated objects, to which them slip according to the vector. Given the isometry character for any points $A$ and $B$ the following identity between distances is satisfied:

$$d (A, B) = d (T (A), T (B)) = d (A ', B')$$

or even more: $\overrightarrow{AB}=\overrightarrow{A'B'}$.

Let's observe that the inverse of a translation is $(T_u)^{-1}=T_{-u}$, so this is the translation of the opposite vector.

As final notes in this section about translations, notice that these preserve the figures identical and they preserve also the same position as originals (with position of the figure we do not refer to the same coordinates on the plane).

Finally, we are going to study how we must proceed to calculate the translations of the following figures:

From these three basic transformations, it will be possible to calculate all the translated of any figure since on the plane any object is reduced to a composition of the three elements previously described.

Given the vector $u = (1,3)$, we will calculate the translation for this vector of the two basic vectors of the plane, that is to say, of the vector $i = (1,0)$ and $j = (0,1)$. Then, to calculate the transformed of i and j we are going to use the system of equations associated with the translation. The above mentioned system has as equations:

$$\begin{pmatrix} x'_1 \\ x'_2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} 1 \\ 3 \end{pmatrix} \Rightarrow \left\{ \begin{array}{l} x'_1=x_1 +1 \\ x'_2=x_ 2+3 \end{array} \right. $$

Therefore, observing the system of equations, the transformed of the vector $i$, $i'= (1 + 1,3) = (2,3)$ and the other basic vector $j' = (1,1 +3) = (1,4)$. Furthermore, if we want to calculate the inverse translation associated with the vector $u$, for the result seen before, it is enough to calculate the translation associated with the vector $-u$. Therefore, the equations system will remain as follows:

$$\begin{pmatrix} x'_1 \\ x'_2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} + \begin{pmatrix} -1 \\ -3 \end{pmatrix} $$

Practice exercises