What is the difference between the vector a, b and the rule (x, y) arrow (x+a, y+b)?

They describe exactly the same translation in different languages. The vector names the shift; the rule tells you the arithmetic to perform on each coordinate. If a problem gives you one, you can write the other instantly. Translating by -4, 3 is the same as applying (x, y) arrow (x-4,\ y+3).

How do I know whether two figures are related by a translation and not some other transformation?

Compute the shift from several pairs of corresponding vertices. If every pair gives the same a, b, it is a translation. You can also look at the picture: in a translation the segments joining corresponding points are all parallel and the same length, and the image has the same orientation as the preimage, with no flipping or turning.

Why does a translation preserve angle measure if the proof only shows distances are preserved?

Because a triangle's angles are determined by its side lengths. Once all corresponding distances match, corresponding triangles are congruent by SSS, so corresponding angles are congruent. Any angle in a figure can be captured by a triangle built from three of its points, so angle preservation follows from distance preservation.

Do I have to graph the figure to translate it?

No. Adding the vector components to each vertex is complete and exact, and it works even when coordinates are large or negative. Graphing is still valuable as a check, especially early on, because a picture will immediately reveal a sign error that a table of numbers might hide.

GEOM-4.1

Translations in the Coordinate Plane

Learn how to translate figures in the coordinate plane with vectors and rules, find the rule mapping a preimage to its image, and see why translations are rigid motions.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Translations in the Coordinate Plane, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

Slide a triangle three units right and two units down and something surprising happens: nothing about the triangle itself changes. Its sides are the same length, its angles are the same size, and its vertices are still labeled clockwise in the same order. Only its address on the grid is different. That is a translation, the first of the rigid motions you will study in this unit.

In this lesson you will learn three skills that show up again and again: applying a translation vector or rule to every point of a figure, working backwards from a preimage and image to recover the rule, and explaining in words and with coordinates why a translation preserves distance, angle measure, and orientation. Those explanations matter, because later in the unit congruence itself gets defined in terms of rigid motions.

What a Translation Does and How We Write It

A translation is a transformation that slides every point of a figure the same distance in the same direction. The original figure is the preimage; the result is the image. Points are matched with prime notation:

A

maps to

A'

B

maps to

B'

, and so on.

There are three common ways to describe the same slide, and your teacher may use any of them.

Notation	Example	Meaning
Vector	$\langle 5, -6 \rangle$	move 5 right, 6 down
Coordinate rule	$(x, y) \rightarrow (x+5, y-6)$	add 5 to each $x$ , subtract 6 from each $y$
Function notation	$T_{5,-6}(x, y)$	the translation by the vector $\langle 5, -6 \rangle$

The key word in the definition is every. You do not slide the figure by eyeballing it; you apply the same addition to each vertex, then connect the image vertices in the same order. For polygons that is enough, because segments map to segments.

Signs carry the direction. A positive first component moves right, negative moves left; a positive second component moves up, negative moves down. A very common error is reading

\langle -3, 4 \rangle

as "3 down, 4 right" because that is how you might describe a slide out loud. The vector is always horizontal first, vertical second, exactly like coordinates.

One more useful fact: the segments joining each point to its image, such as

\overline{AA'}

and

\overline{BB'}

, are all parallel and all the same length. That is the geometric fingerprint of a translation, and it is how you can tell a slide from a rotation on a picture with no rule given.

Applying a Translation to a Figure

To translate a polygon, make a small table of vertices rather than working in your head. Suppose quadrilateral

QRST

has vertices

Q(-4, 1)

R(0, 3)

S(2, -2)

, and

T(-3, -4)

, and you translate by

\langle 6, 2 \rangle

, that is,

(x, y) \rightarrow (x+6, y+2)

Preimage	Add $\langle 6, 2 \rangle$	Image
$Q(-4, 1)$	$(-4+6,\ 1+2)$	$Q'(2, 3)$
$R(0, 3)$	$(0+6,\ 3+2)$	$R'(6, 5)$
$S(2, -2)$	$(2+6,\ -2+2)$	$S'(8, 0)$
$T(-3, -4)$	$(-3+6,\ -4+2)$	$T'(3, -2)$

Notice that arithmetic with negative coordinates is where most mistakes happen, not the geometry. Writing the sum out before simplifying, as in the middle column, catches sign slips.

When you graph the image, label the vertices in the same cyclic order as the preimage. If

QRST

reads counterclockwise, then

Q'R'S'T'

must also read counterclockwise. If yours comes out reversed, you have plotted a point incorrectly, because a translation can never flip a figure.

A quick self-check: pick any one side of the preimage and the corresponding side of the image and compare. Here

QR

goes from

(-4,1)

(0,3)

, a run of 4 and a rise of 2. The side

Q'R'

goes from

(2,3)

(6,5)

, also a run of 4 and a rise of 2. Corresponding sides of a translated figure always have identical slope and identical length, so any mismatch signals an arithmetic error.

Finding the Rule That Maps Preimage to Image

The reverse problem gives you a figure and its image and asks for the rule. Use one pair of corresponding points and subtract in the direction of the mapping:

a = x_{\text{image}} - x_{\text{preimage}}, \qquad b = y_{\text{image}} - y_{\text{preimage}}

For example, if

A(-3, 5)

maps to

A'(4, -1)

, then

a = 4 - (-3) = 7

and

b = -1 - 5 = -6

, so the rule is

(x, y) \rightarrow (x+7, y-6)

, or the vector

\langle 7, -6 \rangle

.

The single biggest error here is subtracting backwards, getting

\langle -7, 6 \rangle

. The reliable habit is to say the subtraction out loud as "image minus preimage," in that order, every time. A second safeguard: after you find the rule, test it on a different pair of corresponding points. If the rule works for

A \rightarrow A'

but not for

B \rightarrow B'

, then either you mislabeled corresponding vertices or the transformation is not a translation at all.

That second possibility is worth taking seriously. Not every pair of congruent figures is related by a translation. If the shifts computed from different vertex pairs disagree, or if the image is a mirror version of the preimage, some other rigid motion is involved.

You can also be asked to reverse a translation. The translation that undoes

(x, y) \rightarrow (x+7, y-6)

(x, y) \rightarrow (x-7, y+6)

; negate both components. This comes up when a problem hands you the image and the rule and asks for the preimage. Rather than memorizing a separate procedure, just solve: if

(x+7, y-6) = (2, 2)

, then

x = -5

and

y = 8

Why a Translation Is a Rigid Motion

A rigid motion is a transformation that preserves distance and angle measure. Translations qualify, and coordinates let you prove it rather than just assert it.

Take any two points

P(x_1, y_1)

and

Q(x_2, y_2)

and translate by

\langle a, b \rangle

. Their images are

P'(x_1+a,\ y_1+b)

and

Q'(x_2+a,\ y_2+b)

. Then

P'Q' = \sqrt{\big((x_2+a)-(x_1+a)\big)^2 + \big((y_2+b)-(y_1+b)\big)^2}

The

a

terms cancel and the

b

terms cancel, leaving

P'Q' = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = PQ

So every distance is unchanged. Since all three side lengths of any triangle in the figure are preserved, corresponding triangles are congruent by SSS, which forces corresponding angles to be equal too. Angle measure is therefore preserved as a consequence of distance being preserved.

Translations also preserve orientation: a figure labeled clockwise stays clockwise. Transformations that preserve orientation are sometimes called direct isometries. Reflections, which you meet in the next lesson, are rigid motions that reverse orientation, so orientation is the feature that distinguishes the two.

A few extra properties follow from distance preservation and are worth stating: parallel lines map to parallel lines, a segment maps to a segment of the same length, and in fact every line maps to a line parallel to itself (or to the same line, if the line happens to run in the direction of the vector). What a translation does not preserve is position, and that is the whole point. A common misconception is that "preserves distance" means points do not move. It means the distance between any two points is unchanged, even though both points move.

Key terms

Translation.: A transformation that slides every point of a figure the same distance in the same direction, described by a vector $\langle a, b \rangle$ or the rule $(x, y) \rightarrow (x+a, y+b)$ .
Preimage.: The original figure before a transformation is applied. Its points are usually labeled $A$ , $B$ , $C$ .
Image.: The figure that results from applying a transformation. Its points are labeled with primes, such as $A'$ , $B'$ , $C'$ .
Translation vector.: An ordered pair $\langle a, b \rangle$ giving the horizontal shift $a$ and the vertical shift $b$ of a translation.
Rigid motion (isometry).: A transformation that preserves distance and angle measure, so the image is congruent to the preimage.
Orientation.: The cyclic order (clockwise or counterclockwise) in which the vertices of a figure are labeled. Translations preserve it; reflections reverse it.
Corresponding points.: A point of the preimage and its image under the transformation, such as $B$ and $B'$ ; the segments joining all such pairs in a translation are parallel and congruent.
Component form.: Writing a vector as horizontal change first, vertical change second, matching the order of coordinates.

Worked example

Triangle

ABC

has vertices

A(-3, 4)

B(1, 7)

, and

C(2, -1)

. Translate the triangle by the vector

\langle 5, -6 \rangle

. Then state the rule that maps

\triangle A'B'C'

back onto

\triangle ABC

, and verify that

AB = A'B'

Step 1: Write the rule from the vector. The vector

\langle 5, -6 \rangle

gives

(x, y) \rightarrow (x+5,\ y-6)

.

Step 2: Apply it to each vertex, writing the sums before simplifying.

A(-3, 4) \rightarrow (-3+5,\ 4-6) = A'(2, -2)

B(1, 7) \rightarrow (1+5,\ 7-6) = B'(6, 1)

C(2, -1) \rightarrow (2+5,\ -1-6) = C'(7, -7)

Step 3: Find the rule that undoes the translation. Negate both components:

(x, y) \rightarrow (x-5,\ y+6)

. Check it on one point:

A'(2, -2) \rightarrow (2-5,\ -2+6) = (-3, 4)

, which is

A

. Correct.

Step 4: Verify the distance is preserved.

AB = \sqrt{(1-(-3))^2 + (7-4)^2} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5

A'B' = \sqrt{(6-2)^2 + (1-(-2))^2} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5

AB = A'B' = 5

, as the rigid-motion property predicts.

Step 5: Sanity check on orientation. Going

A \rightarrow B \rightarrow C

and

A' \rightarrow B' \rightarrow C'

traces the vertices in the same rotational direction, so the triangle was slid, not flipped.

Practice questions

Point

M(6, -2)

is the preimage of

M'(-1, 3)

under a translation. Which rule describes the translation?

$(x, y) \rightarrow (x-7,\ y+5)$
$(x, y) \rightarrow (x+7,\ y-5)$
$(x, y) \rightarrow (x-5,\ y+7)$
$(x, y) \rightarrow (x-7,\ y-5)$

Answer: $(x, y) \rightarrow (x-7,\ y+5)$

Subtract image minus preimage in that order:

a = -1 - 6 = -7

and

b = 3 - (-2) = 5

. The rule is

(x, y) \rightarrow (x-7,\ y+5)

. Choosing

(x+7, y-5)

means the subtraction was done backwards, which describes the translation from

M'

M

instead. Choosing

(x-5, y+7)

keeps both signs but swaps the sizes of the two shifts, and

(x-7, y-5)

keeps the correct horizontal shift but moves down instead of up. Always test your rule on the original point:

(6-7,\ -2+5) = (-1, 3)

Segment

\overline{PQ}

has endpoints

P(-2, 5)

and

Q(4, -3)

. It is translated so that the image of

P

P'(3, 1)

. Find the coordinates of

Q'

, and explain how you know

PQ = P'Q'

without computing

P'Q'

from the distance formula.

Answer: $Q'(9, -7)$ ; a translation is a rigid motion, so it preserves distance.

First find the rule from the pair you were given:

a = 3 - (-2) = 5

and

b = 1 - 5 = -4

, so

(x, y) \rightarrow (x+5,\ y-4)

. Apply it to

Q

(4+5,\ -3-4) = Q'(9, -7)

.

For the explanation, use the general argument. When both endpoints have the same numbers

a

and

b

added to their coordinates, those added amounts cancel inside each difference in the distance formula:

(x_2+a)-(x_1+a) = x_2-x_1

and likewise for

y

. Since the differences are unchanged, so is the distance. A translation is therefore a rigid motion, and

PQ = P'Q'

automatically. (If you want the check, both lengths equal

\sqrt{6^2+8^2}=10

A student claims that because a translation "preserves distance," each point of the figure must stay where it is. Explain what is wrong with this reasoning, and describe one visual feature of a translated figure that shows the points really did move.

Answer: Preserving distance refers to distances between pairs of points, not to points staying fixed; every point moves along a segment parallel and congruent to the translation vector.

The property is about the distance from one point of the figure to another, such as

AB

compared with

A'B'

. Every single point moves the same distance in the same direction, so all internal distances stay the same even though the whole figure changes location. Visually, drawing

\overline{AA'}

\overline{BB'}

, and

\overline{CC'}

produces three parallel segments of equal length, which is direct evidence that the points moved and moved together. This is also why a translation with vector

\langle 0, 0 \rangle

is the only one that leaves points fixed.

FAQ

What is the difference between the vector $\langle a, b \rangle$ and the rule $(x, y) \rightarrow (x+a, y+b)$ ?: They describe exactly the same translation in different languages. The vector names the shift; the rule tells you the arithmetic to perform on each coordinate. If a problem gives you one, you can write the other instantly. Translating by $\langle -4, 3 \rangle$ is the same as applying $(x, y) \rightarrow (x-4,\ y+3)$ .
How do I know whether two figures are related by a translation and not some other transformation?: Compute the shift from several pairs of corresponding vertices. If every pair gives the same $\langle a, b \rangle$ , it is a translation. You can also look at the picture: in a translation the segments joining corresponding points are all parallel and the same length, and the image has the same orientation as the preimage, with no flipping or turning.
Why does a translation preserve angle measure if the proof only shows distances are preserved?: Because a triangle's angles are determined by its side lengths. Once all corresponding distances match, corresponding triangles are congruent by SSS, so corresponding angles are congruent. Any angle in a figure can be captured by a triangle built from three of its points, so angle preservation follows from distance preservation.
Do I have to graph the figure to translate it?: No. Adding the vector components to each vertex is complete and exact, and it works even when coordinates are large or negative. Graphing is still valuable as a check, especially early on, because a picture will immediately reveal a sign error that a table of numbers might hide.

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