Mathematics • Year 8 • Unit 3 • Lesson 17
Translations — Mixed Challenge
Pull everything from Lesson 17 together: translating single points and whole shapes, finding the vector from object and image, combining vectors, and reasoning about what stays invariant. Six mixed problems, one "find the mistake", and one open-ended treasure-trail puzzle.
1. Mixed problems — choose the right move
Each question pulls a different idea from Lesson 17. Decide which approach applies before you start writing. Show your working. 3 marks each
1.1 Translate point P(−3, 7) by the column vector (5 over −9). State P′.
1.2 Triangle ABC has A(1, 1), B(4, 1), C(4, 5). Translate the whole triangle by (−2 over 3). List A′, B′, C′.
1.3 Point M(2, 5) maps to M′(6, 1). Find the translation vector, then use it to translate N(0, −3).
1.4 Point R(a, b) is translated by (3 over −5) to land at R′(7, 2). Find the original coordinates a and b.
1.5 A shape is translated first by (3 over −2), then by (−1 over 5). What single vector is equivalent to applying both translations in sequence?
1.6 A translation maps P(−4, 3) to P′(2, −1). Find the vector. Apply the opposite translation (the inverse vector) to bring P′ back to its original. Verify you land at P.
2. Find the mistake
A Year 8 student has tried to find the translation vector that maps A(4, 7) to A′(1, 10). Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — find the vector mapping A(4, 7) to A′(1, 10):
Line 1: Vector = (image − object)
Line 2: Vector = (1 − 4 over 10 − 7)
Line 3: Vector = (3 over 3)
Line 4: So the translation is 3 right and 3 up.
(a) Which line contains the original mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer in column-vector form and in words.
Stuck? Revisit lesson § Card 8 — check the arithmetic 1 − 4. It should be negative.3. Open-ended challenge — Build a treasure trail
This question has more than one valid answer. 4 marks
3.1 A treasure starts at the origin (0, 0) on a coordinate grid. Design a treasure-trail of exactly four consecutive translations that:
- Use whole-number column vectors only (no zeros — every vector has both a non-zero top and bottom).
- End up at the point (10, 6).
- The combined (single equivalent) vector must be (10 over 6).
For your trail:
(i) List the four column vectors in order.
(ii) Plot or list the position after each step (5 positions in total, including start and end).
(iii) Show the check: sum of the four top components = 10, and sum of the four bottom components = 6.
Bonus: at least one of your four vectors must contain a negative component (so the trail "doubles back" at some point).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — P(−3, 7) by (5 over −9)
P′ = (−3 + 5, 7 − 9) = (2, −2).
1.2 — Triangle ABC by (−2 over 3)
A(1, 1) → A′(−1, 4); B(4, 1) → B′(2, 4); C(4, 5) → C′(2, 8). A′(−1, 4), B′(2, 4), C′(2, 8).
1.3 — M(2, 5) → M′(6, 1); then translate N(0, −3)
Vector = (6 − 2 over 1 − 5) = (4 over −4). N′ = (0 + 4, −3 − 4) = (4, −7).
1.4 — R(a, b) by (3 over −5) → R′(7, 2)
a + 3 = 7 → a = 4. b + (−5) = 2 → b = 7. So R = (4, 7). Check: (4 + 3, 7 − 5) = (7, 2) ✓
1.5 — Combined vector
Add components: top = 3 + (−1) = 2; bottom = −2 + 5 = 3. Combined vector = (2 over 3).
1.6 — Vector and inverse
Vector = (2 − (−4) over −1 − 3) = (6 over −4). Inverse = (−6 over 4). Apply to P′(2, −1): (2 − 6, −1 + 4) = (−4, 3) = P ✓
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The student computed 1 − 4 = 3, but 1 − 4 = −3, not 3. They dropped the negative sign in the top component.
(c) Corrected working:
Vector = (image − object) = (1 − 4 over 10 − 7) = (−3 over 3). In words: 3 left and 3 up.
Sanity check: A is at x = 4 and A′ is at x = 1, so the shape moved LEFT, which requires a negative top component.
3 — Treasure trail (sample solution)
There are many valid trails. One good example:
Vectors (in order): (4 over 3), (3 over 2), (1 over 2), (2 over −1).
Positions: (0, 0) → (4, 3) → (7, 5) → (8, 7) → (10, 6). End at (10, 6) ✓
Check: Sum of tops = 4 + 3 + 1 + 2 = 10 ✓. Sum of bottoms = 3 + 2 + 2 + (−1) = 6 ✓. The last vector (2 over −1) has a negative bottom — the trail doubles back downward (bonus condition).
Other valid trails: (5 over 4) + (3 over 3) + (1 over −2) + (1 over 1) = (10 over 6); or (2 over 2) + (5 over 3) + (−1 over 2) + (4 over −1) = (10 over 6). Any four whole-number non-zero vectors that sum to (10 over 6) work.
Marking: 1 mark for four whole-number non-zero vectors; 1 mark for correct intermediate positions; 1 mark for the sum check (10 and 6); 1 bonus mark if at least one vector has a negative component.