Translations

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Learning goals

Describe a translation using corresponding points, displacement arrows, and image notation
Apply a horizontal and vertical change to every point of a figure
Recover a translation from point-image pairs and check the given correspondence
Explain which lengths, angles, and parallel relationships a translation preserves

Follow every point

A motion can be described by telling where each point goes. For example, sliding a triangle $4$ cm to the right sends each corner, each side point, and each interior point $4$ cm to the right. The slide acts on the entire figure.

A transformation is a rule that assigns each point in the plane exactly one image point. The starting figure is the preimage; the resulting figure is the image. Write $P\mapsto P'$ to say that $P$ maps to $P'$ , read “P-prime.” The two are corresponding points, paired by the rule. Thus $A\mapsto A'$ , $B\mapsto B'$ , and $C\mapsto C'$ specify which corners correspond.

Every displacement arrow has the same length and direction, including the arrow for interior point P.

A translation moves every point the same distance in the same direction. It is a slide with no turning, folding, or stretching. A displacement arrow runs from a point to its image. Its length gives the travel distance; its arrowhead gives the direction. All displacement arrows of a nonzero translation have equal lengths and the same direction. They need not lie on the same line.

An arrow $P\to P'$ describes a journey, while the segment $PQ$ joins two points within the figure. Do not confuse how far a point travels with how long a side is. A side can be $7$ cm long even when every point travels $4$ cm.

There is also a zero translation, in which every point stays where it is. Its travel distance is zero, so it has no travel direction. We include it even though an ordinary spoken “slide” suggests movement.

Worked example 1 Translate a segment without a grid

Problem. An arrow from $A$ to $A'$ specifies a translation. It points right and downward and has length $4$ cm. Describe how to draw the image of an oblique segment $UV$ on paper. State how far and in which direction each endpoint travels.

Solution. If $U$ lies on the supporting line through $A$ and $A'$ , use that same line.

If $U$ is off that line, use the compass-and-straightedge parallel-line construction from lesson 2.3 to draw a parallel through the external point $U$ .

On the chosen line, take the ray from $U$ pointing the same way as the given arrow. Use a ruler to mark $U'$ $4$ cm along that ray. Thus $U$ travels $4$ cm in the direction from $A$ to $A'$ .

At $V$ , use the same line-choice method: use the supporting line through $A$ and $A'$ if $V$ lies on it; otherwise, construct the parallel through the external point $V$ .

On that line, choose the ray from $V$ pointing the same way as the given arrow. Use the ruler to mark $V'$ $4$ cm along the ray. The journey from $V$ to $V'$ is also $4$ cm in that direction.

Join $U'$ to $V'$ . Sliding the original segment carries its points onto this image segment. The two copied arrows specify one common move, even though $UV$ is slanted.

Copy one arrow from both endpoints; the segment itself need not point along the travel direction.

Check your understanding

Two points each travel 5 cm. The first moves east, and the second moves north. Could these be the journeys of two points under one translation?

A Yes, equal travel distances are enough B Yes, because both directions are perpendicular C No, the directions must also agree D No, translations can move only one point

Describe the slide with coordinates

Coordinates let you give the same instruction without drawing an arrow. In $(x,y)$ , the first coordinate locates a point horizontally and the second vertically. Positive horizontal change moves right; positive vertical change moves up. Negative changes move left or down.

For example, move $P(2,-1)$ three units right and four units up. Its new horizontal coordinate is $2+3=5$ , and its new vertical coordinate is $-1+4=3$ . The image is $P'(5,3)$ .

Every point receives those same two additions. The rule is

(x,y)\mapsto(x+3,y+4).

The displacement pair $(3,4)$ records the horizontal change and then the vertical change. It is an instruction, not the coordinates of a special point. It also does not say the straight travel distance is $3+4$ : horizontal and vertical counts describe two different directions.

In general, a translation with displacement pair $(a,b)$ has rule

(x,y)\mapsto(x+a,y+b).

The numbers $a$ and $b$ are fixed for the whole plane. They may be positive, negative, or zero. Pair $(0,0)$ gives the zero translation. You already shifted graphs in Algebra I; here the objects are points and plane figures, so the additions act directly on their coordinates.

Worked example 2 Move every vertex of a triangle

Problem. Triangle $PQR$ has $P(1,-2)$ , $Q(4,-1)$ , and $R(2,2)$ . Translate it three units left and two units up. Give the rule and the coordinates of all three image vertices, then draw the image.

Solution. Three left subtracts $3$ from each horizontal coordinate. Two up adds $2$ to each vertical coordinate:

(x,y)\mapsto(x-3,y+2).

For $P$ , calculate each coordinate:

\begin{aligned} x'&=1-3=-2,\\ y'&=-2+2=0. \end{aligned}

Thus $P'=(-2,0)$ .

For $Q$ , the coordinates become $4-3=1$ and $-1+2=1$ , so $Q'=(1,1)$ .

For $R$ , the coordinates become $2-3=-1$ and $2+2=4$ , so $R'=(-1,4)$ .

Plot the three images and connect corresponding sides in the same order, $P'Q'R'$ . Moving only $P$ while leaving $Q$ and $R$ unchanged would change the figure instead of translating it.

The solid preimage and dashed image use the same three-left, two-up instruction at every vertex.

Check your understanding

A translation has displacement pair $(-2,-5)$ . What is the image of $S(-3,4)$ ?

S'(-5,-1)

S'(-1,9)

S'(5,1)

S'(-5,9)

Recover the rule and check the correspondence

To recover a translation, compare an image coordinate with its starting coordinate. Suppose $D(-4,2)$ maps to $D'(1,-1)$ . The horizontal change is $1-(-4)=5$ ; the vertical change is $-1-2=-3$ . The displacement pair is $(5,-3)$ .

This subtraction works because image equals start plus change. Subtracting the start leaves the change. For any specified pair $P(x,y)\mapsto P'(x',y')$ , the changes are $x'-x$ and $y'-y$ .

One point-image pair determines a translation. If other pairs are supplied, each must have those same changes. Agreement checks the listed points; a translation rule then tells where every other point goes. A few matching corners do not establish that an arbitrary unknown transformation acts that way everywhere.

Worked example 3 Check all the named pairs

Problem. A proposed translation sends $A(-2,3)$ to $A'(3,1)$ and $B(1,-1)$ to $B'(6,-3)$ . It is also supposed to send $C(4,2)$ to $C'(8,0)$ . Decide whether one translation can make all three specified matches. If not, give the correct image of $C$ under the translation determined by the first pair.

Solution. The pair for $A$ gives horizontal change $3-(-2)=5$ and vertical change $1-3=-2$ .

The pair for $B$ gives $6-1=5$ and $-3-(-1)=-2$ , so it agrees.

The claimed pair for $C$ gives $8-4=4$ and $0-2=-2$ . Its horizontal change is different. One translation cannot make all three matches.

The first two pairs use $(x,y)\mapsto(x+5,y-2)$ . Applying that rule to $C$ gives

\begin{aligned} x'&=4+5=9,\\ y'&=2-2=0. \end{aligned}

The correct image is $C'(9,0)$ .

Sometimes the image is given and the preimage is missing. For the rule $(x,y)\mapsto(x+2,y-6)$ , an image $(5,-4)$ must have started at $(3,2)$ : its horizontal coordinate needs $2$ added to reach $5$ , and its vertical coordinate needs $6$ subtracted to reach $-4$ . Substitution checks both coordinates.

Check your understanding

A translation sends $A(2,-3)$ to $A'(-1,1)$ . Which proposed second pair fits that same translation?

B(5,2)\mapsto B'(8,6)

B(5,2)\mapsto B'(2,-2)

B(5,2)\mapsto B'(-3,4)

B(5,2)\mapsto B'(2,6)

What stays unchanged?

A slide moves a figure without changing the distances within it or the openings between its rays. Trace an angle measuring $58^\circ$ and a $7$ cm segment on a stiff transparent sheet. Slide the sheet without turning or bending it. The segment has not been pulled longer, and the rays have not opened or closed. The new segment is still $7$ cm long and the new angle is still $58^\circ$ .

Sliding both rays together keeps their opening and every marked segment length unchanged.

We accept this geometric behavior of a slide as a foundation: a translation preserves every distance and every angle measure. Such a motion is called a rigid motion. The tracing gives a picture of that foundation; a few measurements alone would not prove it for every point.

A straight line also slides onto a straight line with the same direction. Two distinct parallel lines therefore remain distinct parallel lines: their common direction and their separation do not change. The image lies on the original line whenever the slide runs along it, since each point then moves along that line. Do not say every image line is distinct from its preimage.

For corresponding points $A\mapsto A'$ and $B\mapsto B'$ , distance preservation says $A'B'=AB$ , equality of lengths. The segments themselves are congruent, $\overline{A'B'}\cong\overline{AB}$ . Likewise, corresponding angles are congruent and their measures are equal. A translation preserves these quantities; their values need not equal the point’s travel distance.

Worked example 4 Use preservation instead of measuring the picture

Problem. A translation maps $D,E,F,G$ to $D',E',F',G'$ respectively. Given $DE=2t+1$ cm, $D'E'=13$ cm, $m\angle DEF=72^\circ$ , and $DE\parallel FG$ , find $t$ and $m\angle D'E'F'$ . State the relationship between $D'E'$ and $F'G'$ .

Solution. The corresponding segment lengths are equal:

\begin{aligned} 2t+1&=13,\\ 2t&=12,\\ t&=6. \end{aligned}

The angle at $E$ corresponds to the angle at $E'$ , so $m\angle D'E'F'=72^\circ$ . This uses angle preservation directly.

The two parallel segments keep their directions under the same slide, so $D'E'\parallel F'G'$ .

Check your understanding

A translation sends triangle ABC to triangle A′B′C′ with the named correspondence. Given $AB=8$ cm and $m\angle BAC=46^\circ$ , which conclusion is guaranteed?

A'B'=8

cm, but the angle may change B

A'B'=8

cm and

m\angle B'A'C'=46^\circ

C Every vertex travels 8 cm D

A'B'=16

cm and

m\angle B'A'C'=46^\circ

The status of the ideas matters when we use them in later reasoning. The earlier accepted foundations and proved angle results remain available from the preceding chapter.

Status	Translation idea	Basis
Accepted foundation	Distance and angle preservation; straight lines keep their direction	The geometric picture of sliding a stiff sheet
Derived result	Add the same horizontal and vertical changes to every coordinate pair	The meaning of coordinates and a common displacement
Derived result	Distinct parallel lines have distinct parallel images	Their directions and separation stay unchanged
Experimental observation	Tracing and measuring a few corresponding points	Evidence for the model, not a proof for all points

Translations

Learning goals

Follow every point

Describe the slide with coordinates

Recover the rule and check the correspondence

What stays unchanged?

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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