Translating 2D shapes
Translating 2D shapes
- Unit 1: Angles & Constructions
- Unit 2: Shapes & Symmetry
- Unit 3: Position & Transformation
- Unit 4: Area, Volume & Symmetry
🎯 In this topic you will
- Work out the coordinates of shapes after a translation
- Translate shapes on a coordinate grid using vectors
🧠 Key Words
- column vector
- congruent
- corresponds
- image
- object
- translate
Show Definitions
- column vector: A two-number vector written vertically, $\begin{bmatrix} a \\ b \end{bmatrix}$, showing a translation of $a$ units in the $x$-direction and $b$ units in the $y$-direction.
- congruent: Shapes that are exactly the same size and shape; one can be moved to fit the other by translations, rotations or reflections.
- corresponds: Describes matching points or sides on an object and its image after a transformation.
- image: The new figure obtained after a transformation of the original shape (the object).
- object: The original shape before any transformation is applied.
- translate: To move a shape without turning or resizing it, by a given vector (same distance and direction for every point).
🔺 Translating a Shape on a Coordinate Grid
You already know that when you translate a 2D shape on a coordinate grid, you move it up or down and right or left.
Look at triangle $ABC$ on this coordinate grid.
The vertices have coordinates $A(1,1)$, $B(2,3)$ and $C(4,1)$.
When you translate triangle $ABC$ 3 squares right and 2 squares up, the vertices now have coordinates $A'(4,3)$, $B'(5,5)$ and $C'(7,3)$.
You say $A'$ as ‘A dash’.
You say that ‘the point $A$corresponds to point $A'$’.
You also say that ‘triangle $ABC$ is the object’ and that ‘triangle $A'B'C'$ is the image of triangle $ABC$’.
The following diagram shows how you can work out the coordinates of the vertices of triangle $A'B'C'$ without drawing a grid.
Translation is 3 squares right $(+3)$ and 2 squares up $(+2)$.
You need to add 3 to all the $x$-coordinates and add 2 to all the $y$-coordinates of the vertices.
| Vertex | Coordinates | Translation | New Coordinates |
|---|---|---|---|
| A | $(1,1)$ | → +3, ↑ +2 | $A'(4,3)$ |
| B | $(2,3)$ | → +3, ↑ +2 | $B'(5,5)$ |
| C | $(4,1)$ | → +3, ↑ +2 | $C'(7,3)$ |
❓ EXERCISES
1. A triangle, ABC, has vertices at the points $A(2,1)$, $B(7,1)$ and $C(2,5)$. ABC is translated 4 squares right and 5 squares up. The image of ABC is $A'B'C'$. Copy and complete the workings to find the coordinates of the vertices of $A'B'C'$.
👀 Show answer
$A(2,1) \;\to\; A'(6,6)$
$B(7,1) \;\to\; B'(11,6)$
$C(2,5) \;\to\; C'(6,10)$
2. The yellow cards have different translations written on them. The white cards show what must be added or subtracted to the $x$- and $y$-coordinates of a shape to complete the translation. Match each yellow card to its correct white card. The first one has been done for you.
👀 Show answer
B → ii ($+4,\;-1$)
C → iii ($-4,\;-1$)
D → iv ($+4,\;+1$)
❓ EXERCISES
3. A parallelogram, $PQRS$, has vertices at the points $P(3,1)$, $Q(8,1)$, $R(10,4)$ and $S(5,4)$.
$PQRS$ is translated $1$ square left and $3$ squares up. The image of $PQRS$ is $P'Q'R'S'$.
Work out the coordinates of the vertices of $P'Q'R'S'$.
👀 Show answer
$P(3,1) \to P'(2,4)$ $Q(8,1) \to Q'(7,4)$ $R(10,4) \to R'(9,7)$ $S(5,4) \to S'(4,7)$
🧠 Think like a Mathematician
4. Work on your own first, then compare with a classmate.
Dan’s homework:
A triangle, $ABC$, has vertices at the points $A(1,4)$, $B(2,7)$ and $C(5,6)$.
$ABC$ is translated $6$ squares right and $4$ squares down.
The image of $ABC$ is $A'B'C'$. Work out the coordinates of the vertices of $A'B'C'$.
Dan’s solutions:$A'(7,0)$, $B'(8,11)$, $C'(1,12)$.
a. Dan has correctly worked out the coordinates of only one vertex. Which is the correct vertex: $A'$, $B'$ or $C'$?
b. Explain the mistakes Dan made when he worked out the other vertices.
c. What could Dan do to improve the solution that he has written?
d. How could he check that his answers are correct?
👀 show answer
- a. The correct vertex is $A'(7,0)$. Translation by $\begin{bmatrix} +6 \\ -4 \end{bmatrix}$ maps $A(1,4)$ to $(1+6,\ 4-4)=(7,0)$.
- b. Dan added $4$ instead of subtracting $4$ to the $y$-coordinates, and for $C$ he also moved left instead of right.
Correct mappings: $B(2,7)\mapsto (2+6,\ 7-4)=(8,3)$ (not $(8,11)$), $C(5,6)\mapsto (5+6,\ 6-4)=(11,2)$ (not $(1,12)$). - c. Write and apply the rule explicitly: $(x,y)\mapsto(x+6,\ y-4)$ to each vertex; or use the column vector $\begin{bmatrix}6\\-4\end{bmatrix}$ and show working in a small table.
- d. Plot both triangles on a coordinate grid: corresponding vertices should be in the same position relative to each other, and the segments $AA'$, $BB'$, $CC'$ should be parallel and equal, each with vector $\begin{bmatrix}6\\-4\end{bmatrix}$. Alternatively, check by reversing the translation with $\begin{bmatrix}-6\\+4\end{bmatrix}$ to recover the original points.
❓ EXERCISES
5. A pentagon, $JKLMN$, has vertices at $J(1,3)$, $K(3,3)$, $L(3,5)$, $M(2,7)$ and $N(1,5)$. The pentagon is translated $3$ squares right and $2$ squares down to become $J'K'L'M'N'$.
a. Work out the coordinates of the vertices of $J'K'L'M'N'$.
b. i. On a square grid, draw some coordinate axes going from $0$ to $8$ on the $x$- and $y$-axes.
ii. Draw the pentagon $JKLMN$ on the grid.
iii. Translate the pentagon $3$ squares right and $2$ squares down to become $J'K'L'M'N'$.
iv. Use your grid to check that your coordinates in part $a$ for the vertices of $J'K'L'M'N'$ are correct. If they are incorrect, make sure you understand the mistakes that you have made.
👀 Show answer
a. Translation vector $\begin{bmatrix}+3\\-2\end{bmatrix}$ (add $3$ to all $x$-coordinates; subtract $2$ from all $y$-coordinates):
$J(1,3)\to J'(4,1)$, $K(3,3)\to K'(6,1)$, $L(3,5)\to L'(6,3)$, $M(2,7)\to M'(5,5)$, $N(1,5)\to N'(4,3)$.
b. i. Draw axes from $0$ to $8$ on both axes (equal square scale).
ii. Plot $J(1,3)$, $K(3,3)$, $L(3,5)$, $M(2,7)$, $N(1,5)$ and join in order.
iii. Move each vertex by $\begin{bmatrix}+3\\-2\end{bmatrix}$ to get $J'K'L'M'N'$.
iv. Read coordinates of the image from the grid and confirm they match part $a$: $J'(4,1)$, $K'(6,1)$, $L'(6,3)$, $M'(5,5)$, $N'(4,3)$.
❓ EXERCISES
6. Chaow translates triangle $JKL$ to $J'K'L'$. $JKL$ has vertices at $J(4,2)$, $K(5,5)$ and $L(3,3)$. Chaow works out that the vertices of $J'K'L'$ are at $J'(1,7)$, $K'(10,2)$ and $L'(0,8)$. Chaow has worked out two of the vertices correctly and one incorrectly.
a. Which vertex, $J'$, $K'$ or $L'$, is incorrect? Explain how you worked out your answer.
b. What is the correct translation that Chaow used?
c. What is the incorrect translation that Chaow used?
👀 Show answer
a.$K'$ is incorrect. Check vectors:
- $J(4,2)\to J'(1,7)$ gives $\begin{bmatrix}-3\\+5\end{bmatrix}$.
- $L(3,3)\to L'(0,8)$ gives $\begin{bmatrix}-3\\+5\end{bmatrix}$.
- $K(5,5)\to K'(10,2)$ gives $\begin{bmatrix}+5\\-3\end{bmatrix}$ (does not match).
b. Correct translation: $\begin{bmatrix}-3\\+5\end{bmatrix}$ (left $3$, up $5$).
c. Incorrect translation used for $K'$: $\begin{bmatrix}+5\\-3\end{bmatrix}$ (right $5$, down $3$).
7. This is part of Hathai’s homework. She has spilt some juice on her work.
Question (visible parts): A square, $ABCD$, has vertices at the points $A(-3,-3)$, $B(2,-3)$, $C(\dots)$ and $D(\dots)$. $ABCD$ is translated to $A'B'C'D'$. The answer line shows $A'(-7,3)$, $B'(\dots)$, $C'(-2,2)$, $D'(-7,-2)$.
a. Work out the coordinates of vertices: i.$B'$ii.$C$iii.$D$
b. Explain how you worked out the answers to part $a$.
👀 Show answer
a. i. From $A(-3,-3)$ to $A'(-7,3)$ the translation is $\begin{bmatrix}-4\\+6\end{bmatrix}$. Apply it to $B(2,-3)$: $B'=(2-4,\ -3+6)=(-2,3)$.
a. ii. Since $C'(-2,2)=C+\begin{bmatrix}-4\\+6\end{bmatrix}$, $C=(-2+4,\ 2-6)=(2,-4)$.
a. iii. Since $D'(-7,-2)=D+\begin{bmatrix}-4\\+6\end{bmatrix}$, $D=(-7+4,\ -2-6)=(-3,-8)$.
b. Find the translation from the known pair $A\to A'$: subtract $4$ from all $x$-coordinates and add $6$ to all $y$-coordinates (vector $\begin{bmatrix}-4\\+6\end{bmatrix}$). Apply this to get $B'$, and reverse it (add $4$ to $x$, subtract $6$ from $y$) to recover the original $C$ and $D$ from their images.
🔁 What is a translation?
You already know that when you translate a $2$D shape on a coordinate grid, you move it up or down and right or left.
📦 Describing the movement
You can describe this movement with a column vector.
🧮 Example of a column vector
This is an example of a column vector: $\begin{bmatrix} 2 \\ 5 \end{bmatrix}$.
↔️ What the top number means
The top number tells you how many units to move the shape right (positive number) or left (negative number).
↕️ What the bottom number means
The bottom number tells you how many units to move the shape up (positive number) or down (negative number).
🧭 Reading vectors
For example: $\begin{bmatrix} 2 \\ 5 \end{bmatrix}$ means ‘move the shape $2$ units right and $5$ units up’.
$\begin{bmatrix} -2 \\ -3 \end{bmatrix}$ means ‘move the shape $2$ units left and $3$ units down’.
📏 Squares and units
If the scale on the grid is one square to one unit, the numbers tell you how many squares to move the object up/down and across.
✅ What translations preserve
When a shape is translated, only its position changes. Its shape and size stay the same. This means that the object and its image are always congruent.
❓ EXERCISES
1. The yellow cards show translations.
The blue cards show column vectors.
Match each yellow card with the correct blue card.
The first one is done for you: A and iii.
👀 Show answer
B → i ($\begin{bmatrix}4 \\ -1\end{bmatrix}$)
C → iv ($\begin{bmatrix}-4 \\ -1\end{bmatrix}$)
D → ii ($\begin{bmatrix}4 \\ 1\end{bmatrix}$)
❓ EXERCISES
2. The diagram shows triangle P on a coordinate grid.
Copy the grid, then draw the image of triangle P after each translation.
a. $\begin{pmatrix}3 \\ 2\end{pmatrix}$
b. $\begin{pmatrix}2 \\ -2\end{pmatrix}$
c. $\begin{pmatrix}-1 \\ 3\end{pmatrix}$
d. $\begin{pmatrix}-2 \\ -1\end{pmatrix}$
👀 Show answer
b. Move triangle $P$ $2$ squares right and $2$ squares down.
c. Move triangle $P$ $1$ square left and $3$ squares up.
d. Move triangle $P$ $2$ squares left and $1$ square down.
3. The diagram shows shape A on a coordinate grid.
Copy the grid, then draw the image of shape A after each translation.
a. $\begin{pmatrix}3 \\ 2\end{pmatrix}$
b. $\begin{pmatrix}4 \\ -2\end{pmatrix}$
c. $\begin{pmatrix}-2 \\ 2\end{pmatrix}$
d. $\begin{pmatrix}-1 \\ -2\end{pmatrix}$
👀 Show answer
b. Move shape $A$ $4$ squares right and $2$ squares down.
c. Move shape $A$ $2$ squares left and $2$ squares up.
d. Move shape $A$ $1$ square left and $2$ squares down.
4. This is part of Fin’s homework.
a. Is Fin’s answer correct? Explain your answer.
b. How could Fin check whether his answer is correct?
👀 Show answer
b. Fin could check his answer by applying the translation to $A'B'C'$ and seeing if it returns exactly to $ABC$.
🧠 Think like a Mathematician
5. Work on your own, then compare with a partner.
Zara says: “If I translate a shape using the column vector $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$, I can translate the shape back to its original position using the column vector $\begin{pmatrix} -2 \\ -3 \end{pmatrix}$.”
a. Show that Zara is correct.
b. Write the column vectors that translate a shape back to its original position after these translations:
c. When a shape is translated using a column vector, it can be translated back to its original position. Write a general rule for finding the column vector that will translate a shape back to its original position.
Show Answers
- a. Translating by $\begin{pmatrix}2\\3\end{pmatrix}$ sends $(x,y)$ to $(x+2,\ y+3)$. Translating by $\begin{pmatrix}-2\\-3\end{pmatrix}$ then gives $(x+2-2,\ y+3-3)=(x,y)$. So Zara is correct.
- b. The “undo” vector is the negative of the given vector:
- i. Back vector: $\begin{pmatrix}4\\-7\end{pmatrix}$
- ii. Back vector: $\begin{pmatrix}-3\\5\end{pmatrix}$
- iii. Back vector: $\begin{pmatrix}2\\8\end{pmatrix}$
- c. If the translation vector is $\begin{pmatrix}a\\b\end{pmatrix}$, then the vector that returns the shape to its original position is $\begin{pmatrix}-a\\-b\end{pmatrix}$ (add the negatives of both components).
❓ EXERCISES
6. The diagram shows triangle $DEF$.
a. Copy the grid, then draw the image of the triangle after the translation $\begin{bmatrix} 3 \\ -2 \end{bmatrix}$.
Label the triangle $D'E'F'$.
b. Copy and complete these statements.
$\angle D'E'F' = \ \ \ \ \ ^\circ$, $\angle D'F'E' = \ \ \ \ \ ^\circ$ and $\angle E'D'F' = \ \ \ \ \ ^\circ$.
$D'F'$ has a length of ...... units.
c. Copy and complete these statements.
Choose from the words in the box.
When you compare an object and its image after a translation:
• corresponding lengths are ...............
• corresponding angles are ...............
• the object and the image are ...............
👀 Show answer
a. Translate every vertex by $\begin{bmatrix} 3 \\ -2 \end{bmatrix}$ (right $3$, down $2$) to get $D'E'F'$.
b. Translations preserve angles and lengths, so $\angle D'E'F' = 90^\circ$, $\angle D'F'E' = 45^\circ$, $\angle E'D'F' = 45^\circ$; $D'F' = 4$ units.
c. corresponding lengths are equal; corresponding angles are equal; the object and the image are congruent.
7. The diagram shows two shapes, $P$ and $Q$.
Choose the column vector (A, B or C) that translates
a. shape $P$ to shape $Q$ A $\begin{bmatrix} 2 \\ 3 \end{bmatrix}$ B $\begin{bmatrix} -2 \\ 3 \end{bmatrix}$ C $\begin{bmatrix} 2 \\ -3 \end{bmatrix}$
b. shape $Q$ to shape $P$ A $\begin{bmatrix} 2 \\ 3 \end{bmatrix}$ B $\begin{bmatrix} -2 \\ 3 \end{bmatrix}$ C $\begin{bmatrix} 2 \\ -3 \end{bmatrix}$
👀 Show answer
a.C — move right $2$, down $3$ ($\begin{bmatrix} 2 \\ -3 \end{bmatrix}$).
b.B — the reverse vector: left $2$, up $3$ ($\begin{bmatrix} -2 \\ 3 \end{bmatrix}$).
❓ EXERCISES
8. The diagram shows shapes $L, M, N, P$ and $Q$ on a coordinate grid. Write the column vector that translates
a. shape $N$ to shape $L$
b. shape $N$ to shape $P$
c. shape $N$ to shape $Q$
d. shape $N$ to shape $M$
e. shape $L$ to shape $P$
f. shape $P$ to shape $M$
👀 Show answer
Reading the grid (matching like-for-like points on each identical shape):
a.$N \to L:\ \begin{bmatrix} 4 \\ 2 \end{bmatrix}$
b.$N \to P:\ \begin{bmatrix} 3 \\ -2 \end{bmatrix}$
c.$N \to Q:\ \begin{bmatrix} 0 \\ -3 \end{bmatrix}$
d.$N \to M:\ \begin{bmatrix} -2 \\ 1 \end{bmatrix}$
e.$L \to P:\ \begin{bmatrix} -1 \\ -4 \end{bmatrix}$
f.$P \to M:\ \begin{bmatrix} -5 \\ 3 \end{bmatrix}$
🧠 Think like a Mathematician
9. The diagram shows triangle $JKL$. Marcus and Arun translate triangle $JKL$ using the column vector $\begin{pmatrix}5\\-4\end{pmatrix}$. They label the image $J'K'L'$. Read what Marcus and Arun say, then answer the questions.
a. Explain why Marcus is correct and Arun is incorrect.
b. Use Marcus’s method to calculate the coordinates of $K'$ and $L'$.
c. Use the diagram to check that your answers are correct. Discuss your methods and answers to parts $a$ and $b$ with other learners in your class.
Show Answers
- a. A translation by $\begin{pmatrix}5\\-4\end{pmatrix}$ sends $(x,y)\mapsto(x+5,\ y-4)$. Marcus adds the components correctly: for $J(-3,4)$, $J'=(-3+5,\ 4-4)=(2,0)$. Arun’s working ends with a column vector$\begin{pmatrix}2\\0\end{pmatrix}$, which is not a coordinate pair; the image point should be written as the ordered pair $(2,0)$. So Marcus is correct; Arun’s form is incorrect.
- b. Read the original base points from the diagram: $K(-4,2)$ and $L(-2,2)$. Apply the vector $\begin{pmatrix}5\\-4\end{pmatrix}$ component-wise:
$K' = (-4+5,\ 2-4)=(1,-2)$
$L' = (-2+5,\ 2-4)=(3,-2)$ - c. On the grid, check that the three displacements are identical: $\overrightarrow{JJ'}=\overrightarrow{KK'}=\overrightarrow{LL'}=\begin{pmatrix}5\\-4\end{pmatrix}$. Also, the image triangle $J'K'L'$ is congruent to $JKL$ and parallel corresponding sides confirm the translation.
❓ EXERCISES
10. A rectangle $ABCD$ has vertices at the points $A(-2,3)$, $B(4,3)$, $C(4,-2)$, $D(-2,-2)$.
$ABCD$ is translated using the column vector $\begin{bmatrix} 8 \\ 5 \end{bmatrix}$.
a. Calculate the coordinates of $A', B', C'$ and $D'$.
b. Check your answers are correct by drawing a diagram and translating rectangle $ABCD$.
c. Compare and discuss your working for part $a$ with that of a partner. Have you used the same methods? Are both sets of working easy to understand?
👀 Show answer
a. Add $8$ to each $x$ and $5$ to each $y$:
$A'(-2+8,\ 3+5)=(6,8)$, $B'(4+8,\ 3+5)=(12,8)$, $C'(4+8,\ -2+5)=(12,3)$, $D'(-2+8,\ -2+5)=(6,3)$.
b. Plot $ABCD$, then move every vertex by $\begin{bmatrix}8\\5\end{bmatrix}$; the image should be a congruent rectangle with vertices as in part $a$.
c. Valid methods include a table of additions or vector notation. Clear working shows component-wise addition to each vertex.
11. This is part of Joule’s classwork. She has spilt some juice on her work.
a. Work out the coordinates of vertices i.$F'$ ii.$G$ iii.$H$
b. Explain how you worked out the answers to part $a$.
👀 Show answer
The translation from $E(-5,-1)$ to $E'(-8,6)$ is the vector $\begin{bmatrix}-3\\7\end{bmatrix}$.
a. i.$F(3,-1)$ → $F'=(3-3,\ -1+7)=(0,6)$.
a. ii. From $G'(0,14)$, recover $G$ by reversing the vector: $G=(0+3,\ 14-7)=(3,7)$.
a. iii. From $H'(-8,14)$, $H=(-8+3,\ 14-7)=(-5,7)$.
b. Found the column vector using the known pair $E \to E'$. Applied the same vector to get $F'$; added the inverse vector $\begin{bmatrix}3\\-7\end{bmatrix}$ to images to recover the originals $G$ and $H$.