Standards: G-CO.A.2 · G-CO.A.4 · G-CO.A.5 · G-CO.B.6
Marisol is moving next weekend. She draws her new living room on graph paper, one square per square foot, then cuts a scrap of paper into a rectangle the size of the couch and slides it around the plan.
She moves the cutout 4 squares right and 3 squares down. Every corner of it moved that same 4 and 3. The couch did not spin, flip, or change size.
If one corner started at (2, 5), where does it land?
Why paper, and why it matters
The cutout does the work a rule will do. Marisol moves one piece of paper, but all four corners move by the identical amount — because they are attached to each other.
Write the corners as coordinates and you can find where they land without the scissors. That is the difference between a couch and a whole floor of desks.
Given a point or figure and a translation vector, you will produce the image, recover a vector from corresponding points, and check that every point uses one displacement.
Definitions and rules
Translation by the vector ⟨a, b⟩
a shifts horizontally — positive right, negative left.
b shifts vertically — positive up, negative down.
Why it has to be this: one displacement applies to every point, and horizontal movement only touches x.
Rebuild the rule from scratch
Three facts. Everything else follows.
So if the displacement is a across and b up, fact 3 gives x → x + a and y → y + b, and fact 1 says the same a and b apply to every point.
No other rule fits all three. If you ever forget whether it is plus or minus, rebuild it from fact 1.
A transformation that preserves distance is called a rigid motion, and its image is congruent to the preimage.
For a translation you can prove it in one line rather than measuring: the displacement cancels.
What that proves
Distance between two points depends only on the difference of their coordinates. Both differences come out unchanged, so the distance cannot change either.
That holds for any figure and any vector — not just one triangle.
Worked examples
Example 1 — recovering the vector
M(1, 5) maps to M′(4, 1). Find the vector.
Suggested work
Givens
Tools
Primitives
Computation
Notice: the Tools row changed. Same layout, different tool — choosing the right one is part of the work.
Example 2 — a whole figure
Triangle ABC has vertices A(−3, 4), B(1, −2), C(5, 3). Apply T⟨−4, −6⟩.
Suggested work
Givens
Tools
Primitives
Computation
Notice: the parameters are listed once; only the inputs repeat.
Why they cannot change
One displacement, every point. The parameters are fixed before you start — that is what makes it a translation rather than three separate moves.
It is also why eyeballing the second vertex fails: you would be guessing at something already decided.
Example 3 — one point with positive components
Apply T⟨3, 4⟩ to J(2, 1).
Suggested work
Givens
Tools
Primitives
Computation
Notice: The horizontal component changes x, and the vertical component changes y.
Why one rule handles both coordinates
A translation uses one displacement. Horizontal movement changes only x, and vertical movement changes only y.
Common traps
Common trap
Moving the first vertex correctly, then eyeballing the rest. Write the arithmetic for every vertex — especially with negatives, where “down 6” and “+ (−6)” are the same move but look different on the page.
Not the same thing
Every translation is a rigid motion, but not every rigid motion is a translation. Rotate (1, 0) a quarter turn about the origin and it moves by (−1, 1); rotate (3, 0) and it moves by (−3, 3). Different points, different displacements — so the one-displacement rule fails. Rigid, but not a translation.
Practice
Apply T⟨−5, 0⟩ to E(2, −4).
Answer — tap to show
E′(−3, −4)
Suggested work — tap to show
Givens
E(2, −4) T⟨−5, 0⟩Tools
T⟨a, b⟩(x, y) = (x + a, y + b)Primitives
x = 2 y = −4 a = −5 b = 0Computation
= ((2) + (−5), (−4) + (0)) = (−3, −4) b = 0, so y does not move at all.Apply T⟨3, −2⟩ to D(−1, 6).
Answer — tap to show
D′(2, 4)
Suggested work — tap to show
Givens
D(−1, 6) T⟨3, −2⟩Tools
T⟨a, b⟩(x, y) = (x + a, y + b)Primitives
x = −1 y = 6 a = 3 b = −2Computation
T⟨a, b⟩(x, y) = (x + a, y + b) = ((−1) + (3), (6) + (−2)) = (2, 4)A(−2, 3) maps to A′(4, 1), and B(1, −1) maps to B′(7, −3). Decide whether one translation maps both points, and name it.
Answer — tap to show
Yes. Both points move by ⟨6, −2⟩, so the translation is T⟨6, −2⟩.
Suggested work — tap to show
Givens
A(−2, 3) → A′(4, 1) B(1, −1) → B′(7, −3)Tools
a = x′ − x b = y′ − yPrimitives
Check the displacement for each point.Computation
A: a = 4 − (−2) = 6, b = 1 − 3 = −2 B: a = 7 − 1 = 6, b = −3 − (−1) = −2 The vectors match: T⟨6, −2⟩.A classmate applies T⟨0, 7⟩ to F(−3, −3) and writes F′(4, −3). Find their mistake.
Answer — tap to show
They are wrong. The correct image is F′(−3, 4).
Suggested work — tap to show
Givens
F(−3, −3) T⟨0, 7⟩ Classmate wrote F′(4, −3)Tools
T⟨a, b⟩(x, y) = (x + a, y + b)Primitives
x = −3 y = −3 a = 0 b = 7Computation
= ((−3) + (0), (−3) + (7)) = (−3, 4) The mistake: they applied b = 7 to x instead of y. That gives ((−3) + (7), (−3) + (0)) = (4, −3), which is what they wrote.K(−2, 3) maps to K′(6, −1). Find the vector.
Answer — tap to show
T⟨8, −4⟩
Suggested work — tap to show
Givens
K(−2, 3) K′(6, −1)Tools
a = x′ − x b = y′ − yPrimitives
x = −2 y = 3 x′ = 6 y′ = −1Computation
a = (6) − (−2) = 8 b = (−1) − (3) = −4 Rule: T⟨8, −4⟩ Check: ((−2) + (8), (3) + (−4)) = (6, −1) ✓
Try it
Translation explorer
T⟨3, −2⟩(x, y) = (x + 3, y − 2)
| Vertex | Preimage | Addition | Image |
|---|