Mathematics • Year 8 • Unit 3 • Lesson 17
Translations
Build fluency with the column-vector rule (x, y) → (x + a, y + b). One fully worked example, one guided example with blanks, then eight independent problems ramping from single-point shifts to finding the vector from object and image.
1. I do — fully worked example
Read every line. Each step has a short reason so you can see why we do it, not just what.
Problem. Triangle ABC has vertices A(1, 0), B(3, 0), C(2, 3). Translate it by the column vector (2 over −1). Find A′, B′, C′.
Step 1 — Read the vector.
Top = 2 (move right 2). Bottom = −1 (move down 1).
Reason: in (a over b), top is horizontal shift and bottom is vertical shift. Negative bottom = down.
Step 2 — Write the rule.
(x, y) → (x + 2, y − 1)
Reason: add the top to x; add the bottom to y.
Step 3 — Apply to each vertex.
A(1, 0) → A′(1 + 2, 0 − 1) = A′(3, −1)
B(3, 0) → B′(3 + 2, 0 − 1) = B′(5, −1)
C(2, 3) → C′(2 + 2, 3 − 1) = C′(4, 2)
Reason: the same translation applies to every vertex, not just one.
Step 4 — Check.
Each vertex has moved exactly 2 right and 1 down — the shape is the same size and shape, just in a new position.
Answer: A′(3, −1), B′(5, −1), C′(4, 2)
2. We do — fill in the missing steps
Same shape as Section 1, with the working faded. Fill in each blank. 4 marks
Problem. Triangle KLM has vertices K(0, 1), L(4, 1), M(2, 5). Translate it by the column vector (−3 over −4). Find K′, L′, M′.
Step 1 — Read the vector: top = ______ (move left/right by ______), bottom = ______ (move up/down by ______).
Step 2 — Write the rule:
(x, y) → (x + ______, y + ______)
Step 3 — Apply to each vertex:
K(0, 1) → K′(______, ______)
L(4, 1) → L′(______, ______)
M(2, 5) → M′(______, ______)
Step 4 — Check: every vertex has moved 3 units ______ and 4 units ______.
3. You do — independent practice
Show all working. The first four are foundation (translate one point). The middle two are standard (translate a whole shape). The last two are extension (find the vector from object and image).
Foundation — translate a single point
3.1 Translate P(2, 3) by (5 over 0). State P′. 1 mark
3.2 Translate Q(−1, 4) by (3 over −2). State Q′. 1 mark
3.3 Translate R(0, 0) by (−4 over 6). State R′. 1 mark
3.4 Translate S(5, −3) by (−2 over −1). State S′. 1 mark
Standard — translate a whole shape
3.5 Triangle DEF has D(2, 1), E(6, 1), F(4, 5). Translate by (−1 over 3). List D′, E′, F′. 2 marks
3.6 Square WXYZ has W(0, 0), X(3, 0), Y(3, 3), Z(0, 3). Translate by (4 over −2). List the image vertices. 2 marks
Extension — find the vector (image minus object)
3.7 Point A(2, 4) is translated to A′(5, 1). Find the translation vector. 2 marks
3.8 Point B(−2, 5) is translated to B′(3, −1). Find the translation vector, then use the same vector to translate C(4, 0). 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (vector −3 over −4)
Step 1: top = −3 (move left by 3), bottom = −4 (move down by 4).
Step 2: (x, y) → (x + (−3), y + (−4)).
Step 3: K(0, 1) → K′(−3, −3); L(4, 1) → L′(1, −3); M(2, 5) → M′(−1, 1).
Step 4: each vertex moves 3 units left and 4 units down.
3.1 — P(2, 3) by (5 over 0)
P′ = (2 + 5, 3 + 0) = (7, 3).
3.2 — Q(−1, 4) by (3 over −2)
Q′ = (−1 + 3, 4 − 2) = (2, 2).
3.3 — R(0, 0) by (−4 over 6)
R′ = (0 − 4, 0 + 6) = (−4, 6).
3.4 — S(5, −3) by (−2 over −1)
S′ = (5 − 2, −3 − 1) = (3, −4).
3.5 — Triangle DEF by (−1 over 3)
D(2, 1) → D′(1, 4); E(6, 1) → E′(5, 4); F(4, 5) → F′(3, 8). D′(1, 4), E′(5, 4), F′(3, 8).
3.6 — Square WXYZ by (4 over −2)
W(0, 0) → W′(4, −2); X(3, 0) → X′(7, −2); Y(3, 3) → Y′(7, 1); Z(0, 3) → Z′(4, 1). W′(4,−2), X′(7,−2), Y′(7,1), Z′(4,1).
3.7 — A(2, 4) → A′(5, 1)
Vector = (5 − 2 over 1 − 4) = (3 over −3). Interpretation: 3 right, 3 down.
3.8 — B(−2, 5) → B′(3, −1); then translate C(4, 0)
Vector = (3 − (−2) over −1 − 5) = (5 over −6). Apply to C: C′ = (4 + 5, 0 − 6) = (9, −6).