Translate shapes using column vectors
Translate a whole named polygon by a signed column vector, using one common displacement and separating vector components, movement length and final position.
Section 5.2 · Transformation geometry
Exam scope and assessment notes
These twelve shared lessons apply to Foundation and inherited Higher for F5.2A–M; no extra Higher row is added. Keep named corners, signed turns, equivalent mirror equations, exact fractions, positive factors and the freely chosen-centre and identity exceptions. General-angle coordinates are final3dp approximations of unrounded geometry. Written explanations need human review; coordinate agreement does not certify proof or physical construction. Linear November2026/June2027 entry and regional-centre rules remain separate from modular awards. Book2 Shape and Space7 tags and Hubs66–71 are navigation only: map223–248 differs from proof222/summary247/next248, with chapter/Hub bodies unavailable.
Before you begin
- Read columns as signed horizontal and vertical displacements.
- Add and subtract signed integer and fractional coordinates.
- Distinguish a direct displacement magnitude from an axis-route distance.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What movement does the column[−5;2] specify on ordinary axes?
What you will learn
- Apply both entries of a translation vector to every vertex.
- Describe horizontal and vertical movements with correct signs.
- Distinguish displacement from position and use the opposite vector as the inverse.
Polygon ABC has A(1,1), B(4,1), C(2,3). In the given order, translate by the column vector with top entry 3 and bottom entry -2. Translate the whole polygon by the stated column vector. Give every image vertex, describe its horizontal and vertical displacement, and explain why it is one common movement.
Describe your operation, construct every named image point, and record your actual directed traces. Only the question supplies the givens. The board does not transform a shape or recover a missing parameter for you.
Both Cartesian directions have equal drawing scale. Blue is supplied geometry, amber is my actual work, purple is dirty retained work, and green is a separately gated reference. Screen size is not physical instrument calibration.
Actual boundaries retain their entered order, closure, concavity and wrong or incomplete points. No reflected corner, rotation endpoint, mirror foot or target polygon is filled automatically. Construction rays continue beyond their anchors; mirror lines continue both ways; joining segments keep their original endpoints.
Some actual points are outside this viewport and remain unchanged. The frame clips the drawing, not its mathematical domain. Pan or zoom to inspect them; no artificial arrowhead is placed at a crop intersection.
Supplied translation vector (3, -2). This description does not generate a target boundary.
This finite fixed-case comparison is not proof marking, marks, mastery, physical construction calibration or specialist/formal assessment approval. General-angle points stay unrounded; a separately requested three-decimal written table is an approximation.
Actual operation 1
No actual construction is recorded for this action.
Choose a point slot before adding a keyboard coordinate or picking on the board. Named picks retain exact given or current clean recorded coordinates. Free picks retain their unrounded actual planar coordinates; they do not snap to an expected result.
My written coordinates, quantities and reasoning
Write both coordinates as x,y; exact fractions and finite decimals are supported.
Write both coordinates as x,y; exact fractions and finite decimals are supported.
Write both coordinates as x,y; exact fractions and finite decimals are supported.
Write both coordinates as x,y; exact fractions and finite decimals are supported.
Use a nonnegative exact fraction, decimal or supported single-radical length. Select the stated drawing unit.
Explain your actual correspondence, construction and conditions. A written attempt is not an automatically verified proof.
Explain your actual correspondence, construction and conditions. A written attempt is not an automatically verified proof.
Explain your actual correspondence, construction and conditions. A written attempt is not an automatically verified proof.
This sequence is optional. If you add operations, complete every chosen parameter or remove the sequence. The tool generates no inverse for you.
These original cases support mathematical self-checking. They do not award marks, mastery, proof approval, physical construction competence, specialist signoff or formal assessment readiness. Positive scale factors only; unchanged images do not uniquely recover a centre. Keep general-angle construction coordinates unrounded and separately requested written coordinates at three decimal places.
One displacement acts everywhere
A translation moves every point by the same displacement. With column vector[a;b], add a to each horizontal coordinate and b to each vertical coordinate. The vector specifies a movement, not the image position of one chosen vertex, so different starting positions generally lead to different final coordinates.
Positive and negative entries describe the sense of movement on the stated axes. A zero entry means no movement along that axis, not a missing component. Fractional entries are exact displacements and should be added as fractions rather than rounded or replaced by the nearest grid intersection.
Describe distance and direction without confusing them
The column records both signed perpendicular components. On ordinary right-positive and up-positive axes, a negative first entry means left and a positive second entry means up. If the straight movement length is wanted, use√(a²+b²); adding|a|+|b| would describe a two-leg axis route rather than the single direct translation.
A vector and its negative have equal movement lengths but opposite directions. A zero vector gives the identity translation with no nonzero direction. These statements describe the movement of the whole figure; they do not justify changing the displacement for a vertex that would otherwise leave the visible part of a drawing.
Keep the complete image and verify the inverse
Record every image vertex and join them in the original boundary order. Every corresponding difference P′−P should equal the same column vector. This test is more reliable than checking whether only one side remains parallel or whether the image has roughly the correct size.
The inverse translation uses[−a;−b]. Subtracting the original vector from each image point should return the labelled object. Translation preserves lengths, angles and boundary orientation; a concave corner moves with the rest of the polygon and remains part of the final outline.
Sources for this lesson
- Pearson Mathematics A linear specification · Issue2 · Transformation geometry Foundation printed23/physical29 and Higher inheritance printed38/physical44
- Pearson 2026/2027 International GCSE entry manual · centre regions physical3 and Mathematics A physical25
- Pearson literal second-edition book/Hub map · Book2 Shape and Space7 physical12 · navigation only
Pause and explain
Which vector undoes translation by[7;−4]?
Worked example
Polygon ABC has A(1,1), B(4,1), C(2,3). In the given order, translate by the column vector with top entry 3 and bottom entry -2. Translate the whole polygon by the stated column vector. Give every image vertex, describe its horizontal and vertical displacement, and explain why it is one common movement.
Show the worked solution
- The common movement is column[3;-2]: 3 right and 2 down in coordinate units. Add these signed components to each vertex’s own starting coordinates.
- The complete image is A′=(4,-1), B′=(7,-1), C′=(5,1). Every difference P′−P equals[3;-2], including any zero or fractional component; the column is a movement, not a common final point.
- Reconnect the full named boundary without changing orientation. The direct movement length is √13 coordinate units, while the inverse is translation by column[-3;2]; applying it returns each image point to its labelled object point.
Answer A′=(4,-1), B′=(7,-1), C′=(5,1). Common displacement[3;-2], 3 right and 2 down; direct movement length √13.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Polygon ABCD has A(0,0), B(3,0), C(4,2), D(1,3). In the given order, translate by the column vector with top entry -4 and bottom entry 5. Translate the whole polygon by the stated column vector. Give every image vertex, describe its horizontal and vertical displacement, and explain why it is one common movement.
Give me a hint
Add the same signed column to each starting vertex, not to a single common endpoint.
Compare my reasoning
- The common movement is column[-4;5]: 4 left and 5 up in coordinate units. Add these signed components to each vertex’s own starting coordinates.
- The complete image is A′=(-4,5), B′=(-1,5), C′=(0,7), D′=(-3,8). Every difference P′−P equals[-4;5], including any zero or fractional component; the column is a movement, not a common final point.
- Reconnect the full named boundary without changing orientation. The direct movement length is √41 coordinate units, while the inverse is translation by column[4;-5]; applying it returns each image point to its labelled object point.
A′=(-4,5), B′=(-1,5), C′=(0,7), D′=(-3,8). Common displacement[-4;5], 4 left and 5 up; direct movement length √41.
Look for these in your work
- Apply both entries of a translation vector to every vertex.
- Describe horizontal and vertical movements with correct signs.
- Distinguish displacement from position and use the opposite vector as the inverse.
Polygon ABCDEF has A(0,0), B(4,0), C(4,1), D(1,1), E(1,3), F(0,3). In the given order, translate by the column vector with top entry 0 and bottom entry -3. Translate the whole polygon by the stated column vector. Give every image vertex, describe its horizontal and vertical displacement, and explain why it is one common movement.
Give me a hint
Add the same signed column to each starting vertex, not to a single common endpoint.
Compare my reasoning
- The common movement is column[0;-3]: no horizontal movement and 3 down in coordinate units. Add these signed components to each vertex’s own starting coordinates.
- The complete image is A′=(0,-3), B′=(4,-3), C′=(4,-2), D′=(1,-2), E′=(1,0), F′=(0,0). Every difference P′−P equals[0;-3], including any zero or fractional component; the column is a movement, not a common final point.
- Reconnect the full named boundary without changing orientation. The direct movement length is 3 coordinate units, while the inverse is translation by column[0;3]; applying it returns each image point to its labelled object point.
A′=(0,-3), B′=(4,-3), C′=(4,-2), D′=(1,-2), E′=(1,0), F′=(0,0). Common displacement[0;-3], no horizontal movement and 3 down; direct movement length 3.
Look for these in your work
- Apply both entries of a translation vector to every vertex.
- Describe horizontal and vertical movements with correct signs.
- Distinguish displacement from position and use the opposite vector as the inverse.
Polygon ABCD has A(-2,1), B(1,1), C(2,4), D(-1,3). In the given order, translate by the column vector with top entry -1/2 and bottom entry 3/2. Translate the whole polygon by the stated column vector. Give every image vertex, describe its horizontal and vertical displacement, and explain why it is one common movement.
Give me a hint
Add the same signed column to each starting vertex, not to a single common endpoint.
Compare my reasoning
- The common movement is column[-1/2;3/2]: 1/2 left and 3/2 up in coordinate units. Add these signed components to each vertex’s own starting coordinates.
- The complete image is A′=(-5/2,5/2), B′=(1/2,5/2), C′=(3/2,11/2), D′=(-3/2,9/2). Every difference P′−P equals[-1/2;3/2], including any zero or fractional component; the column is a movement, not a common final point.
- Reconnect the full named boundary without changing orientation. The direct movement length is √(5/2) coordinate units, while the inverse is translation by column[1/2;-3/2]; applying it returns each image point to its labelled object point.
A′=(-5/2,5/2), B′=(1/2,5/2), C′=(3/2,11/2), D′=(-3/2,9/2). Common displacement[-1/2;3/2], 1/2 left and 3/2 up; direct movement length √(5/2).
Look for these in your work
- Apply both entries of a translation vector to every vertex.
- Describe horizontal and vertical movements with correct signs.
- Distinguish displacement from position and use the opposite vector as the inverse.
Common mistakes
- Treating the translation column as a final coordinate.
- Moving vertices by different displacements.
- Dropping a zero or fractional vector component.
What can you explain now?
Can the same translation place two different object vertices at the same image point?
Compare with the explanation
No. Subtracting the same movement from equal images would give equal original positions.
A translation preserves the difference between every pair of vertices, so distinct vertices remain distinct.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Describe one common translation, its inverse and its straight movement length, explaining why the column is not a point location.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.