Math Core

A transformation is a rule that moves every point of the plane to a new location. The simplest one is a translation: every point slides the same distance in the same direction, with no turning and no flipping. In this unit you'll describe transformations precisely with coordinates and vectors, which is exactly what you need later to prove figures congruent or similar.

Pre-images, images and vectors

When a transformation moves a figure, the original is the pre-image and the result is the image. If the pre-image is triangle $ABC$ , the image is written $A'B'C'$ (read "A prime, B prime, C prime"). Each vertex $A$ is sent to its matching vertex $A'$ .

A translation is completely described by how far it moves points horizontally and vertically. Geometry packages those two numbers into a vector, an arrow with a length and a direction. In component form, the vector $\langle a, b \rangle$ means "move $a$ units horizontally and $b$ units vertically."

Definition

Translation

A translation by the vector $\langle a, b \rangle$ moves every point $a$ units horizontally and $b$ units vertically:

(x, y) \rightarrow (x + a,\ y + b)

A positive $a$ moves right and a negative $a$ moves left. A positive $b$ moves up and a negative $b$ moves down.

The angle brackets matter. $(3, -2)$ names a point, a location. $\langle 3, -2 \rangle$ names a vector, a movement. The same vector can be drawn starting from any point.

Worked example: Translating a triangle

Translate triangle $ABC$ with $A(-4, -1)$ , $B(-1, -2)$ and $C(-2, 2)$ by the vector $\langle 5, 2 \rangle$ .

Add $5$ to every x-coordinate and $2$ to every y-coordinate:

\begin{aligned} A(-4, -1) &\rightarrow A'(1, 1) \\ B(-1, -2) &\rightarrow B'(4, 0) \\ C(-2, 2) &\rightarrow C'(3, 4) \end{aligned}

Triangle ABC translated by the vector ⟨5, 2⟩ to triangle A′B′C′.Open in grapher →

Every vertex moved along an arrow of the same length pointing the same way.

Finding the vector

Often you know a point and its image and need the translation. Subtract: the vector from $P(x_1, y_1)$ to $P'(x_2, y_2)$ is

\langle x_2 - x_1,\ y_2 - y_1 \rangle.

Always subtract image minus pre-image. Reversing the order gives the vector that undoes the translation.

Worked example: From one point to the whole rule

A translation maps $P(3, -5)$ to $P'(-1, 2)$ . Write the rule, and find the image of $Q(0, 4)$ .

The vector is $\langle -1 - 3,\ 2 - (-5) \rangle = \langle -4, 7 \rangle$ , so the rule is

(x, y) \rightarrow (x - 4,\ y + 7).

Because a translation moves every point by the same vector, one pair of points is enough to know the whole rule. Then $Q(0, 4) \rightarrow Q'(-4, 11)$ .

Common mistake

Watch the direction of subtraction. If $P(3, -5)$ maps to $P'(-1, 2)$ , the vector is $\langle -4, 7 \rangle$ , not $\langle 4, -7 \rangle$ . Check by adding your vector to $P$ : you should land exactly on $P'$ .

What a translation preserves

Since every point moves identically, a translation can't stretch, bend or flip anything.

Translations are rigid motions

A translation preserves distance and angle measure, so the image is congruent to the pre-image. It also preserves orientation (the vertices $A \to B \to C$ go around in the same direction as $A' \to B' \to C'$ ), and it maps every line to a line parallel to it (or to the same line, if the line points along the vector).

Every point travels the same distance, namely the length of the vector. By the Pythagorean theorem, the vector $\langle a, b \rangle$ has length $\sqrt{a^2 + b^2}$ . Under $\langle 5, 2 \rangle$ in the first example, each vertex moved $\sqrt{25 + 4} = \sqrt{29} \approx 5.4$ units.

Translating a line

You can translate a whole graph, not just a polygon. Suppose you translate the line $y = 2x + 1$ by $\langle 3, 1 \rangle$ . A point $(x, y)$ on the image came from the point $(x - 3,\ y - 1)$ on the original line, so that point must satisfy the original equation.

Worked example: Translating a line

Find an equation of the image of $y = 2x + 1$ under the translation $\langle 3, 1 \rangle$ .

Replace $x$ with $x - 3$ and $y$ with $y - 1$ :

\begin{aligned} y - 1 &= 2(x - 3) + 1 \\ y - 1 &= 2x - 5 \\ y &= 2x - 4 \end{aligned}

The slope is still $2$ , as it should be: a translation maps a line to a parallel line. Check with one point: $(0, 1)$ is on the original line, and its image $(3, 2)$ satisfies $y = 2x - 4$ because $2(3) - 4 = 2$ . ✓

Tip

Translating a graph by $\langle a, b \rangle$ means replacing $x$ with $x - a$ and $y$ with $y - b$ . The minus signs feel backwards, so always test one point of the original graph.

Practice

Practice 1

Translate the point $(-3, 8)$ by the vector $\langle 6, -10 \rangle$ . What is the image?

Enter a point like (2, -3)

Practice 2

Which vector describes the translation that maps $A(2, -1)$ to $A'(-3, 4)$ ?

Practice 3

The translation $(x, y) \rightarrow (x - 4,\ y + 7)$ maps point $P$ to $P'(1, -2)$ . What are the coordinates of $P$ ?

Enter a point like (2, -3)

Practice 4

Every point of a figure is translated by $\langle 6, -8 \rangle$ . How far does each point move?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Triangle $ABC$ has vertices $A(-1, 2)$ , $B(3, 0)$ and $C(1, 5)$ . A translation maps $B$ to $B'(-2, 4)$ . What are the coordinates of $C'$ ?

Enter a point like (2, -3)

Practice 6

A figure is translated by $\langle 3, -2 \rangle$ and then by $\langle -7, 5 \rangle$ . The result is a single translation by $\langle a, b \rangle$ . Enter $a$ and $b$ as an ordered pair $(a, b)$ .

Enter a point like (2, -3)

Practice 7

The line $y = 3x - 2$ is translated by $\langle 2, 1 \rangle$ . Which is an equation of the image?