Translation

Understanding Translation

Translation, also known as a shift or slide, is a type of geometric transformation that moves every point of an object a certain distance in a specified direction. This transformation does not change the orientation, size, or shape of the object; only its position changes.

Definition of Translation

Given any point $P(x,y)$. The translation associated with the vector $\begin{pmatrix} a \\ b \end{pmatrix}$ for point $P(x,y)$, written as $T_{(a,b)}(x,y)$ or $\tau_{(a,b)}(x,y)$, is defined as:

This means:

Here, $a$ is the horizontal shift (positive to the right, negative to the left) and $b$ is the vertical shift (positive upwards, negative downwards).

Translating a Point

A point $(3,2)$ is translated by the vector $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$. Determine the image point of this translation.

Here, $x=3$, $y=2$, $a=-2$, and $b=3$.

Using the formula:

Thus, the image of point $(3,2)$ is $(1,5)$.

Translating a Line

Determine the image of the line $l \equiv 2x + 3y - 1 = 0$ translated by the vector $\begin{pmatrix} -1 \\ 1 \end{pmatrix}$.

Let $P(x,y)$ be any point on line $l$. If translated by the vector $\begin{pmatrix} -1 \\ 1 \end{pmatrix}$, its image is $P'(x',y')$ where:

Substitute these values of $x$ and $y$ into the equation of line $l$:

Replacing $x'$ and $y'$ back to $x$ and $y$, the equation of the image line $l'$ is:

Exercises

1. A point $(-4,4)$ is translated by the vector $\begin{pmatrix} -2 \\ -3 \end{pmatrix}$. Determine the image point of this translation.
2. Determine the image of the line $l' \equiv 5x - 2y + 3 = 0$ translated by the vector $\begin{pmatrix} 2 \\ -1 \end{pmatrix}$.
3. A triangle with vertices $A(1,1)$, $B(4,1)$, and $C(1,5)$ is translated by the vector $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$. Determine the coordinates of the image triangle $A'B'C'$!

Key Answers

1. Point $(-4,4)$, vector $\begin{pmatrix} -2 \\ -3 \end{pmatrix}$. $x=-4, y=4, a=-2, b=-3$.

   $$x' = -4 + (-2) = -6$$

   $$y' = 4 + (-3) = 1$$

   Thus, the image point is $(-6,1)$.

2. Line $5x - 2y + 3 = 0$, vector $\begin{pmatrix} 2 \\ -1 \end{pmatrix}$. $a=2, b=-1$.

   $$x' = x + 2 \implies x = x' - 2$$

   $$y' = y - 1 \implies y = y' + 1$$

   Substitute into the line equation:

   $$5(x' - 2) - 2(y' + 1) + 3 = 0$$

   $$5x' - 10 - 2y' - 2 + 3 = 0$$

   $$5x' - 2y' - 9 = 0$$

   Image line equation: $5x - 2y - 9 = 0$.

3. Points $A(1,1)$, $B(4,1)$, $C(1,5)$. Vector $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$.

   $$A'(1+2, 1+3) = A'(3,4)$$

   $$B'(4+2, 1+3) = B'(6,4)$$

   $$C'(1+2, 5+3) = C'(3,8)$$

   The coordinates of the image triangle are $A'(3,4)$, $B'(6,4)$, and $C'(3,8)$.