Find vector components from initial and terminal coordinates
Subtract initial coordinates from terminal coordinates · Reverse an arrow by reversing the difference · Check a displacement by adding it to its starting point
Section Paper 2 · Historical Extended · non-calculator
Exam scope and assessment notes
Extended content; not a Core requirement. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Directed columns
- Signed subtraction
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Subtract initial coordinates from terminal coordinates
- Reverse an arrow by reversing the difference
- Check a displacement by adding it to its starting point
End minus start
For A(x₁,y₁) and B(x₂,y₂), →AB = [x₂−x₁; y₂−y₁]. Keep the same order in both components. For A(−2,1), B(3,−3), this gives [5; −4]. Reverse direction gives →BA = [−5; 4].
Verify the endpoint
Add the displacement to A to recover B: (−2+5,1−4) = (3,−3). This check detects a reversed subtraction or a lost negative sign. A displacement vector is independent of the coordinate origin, even though the two position coordinates change when the origin moves.
Sources for this lesson
Pause and explain
Which subtraction gives →AB?
Worked example
Find →AB for A(−2,1) and B(3,−3), and verify the end.
Show the worked solution
- Horizontal component 3−(−2) = 5; vertical −3−1 = −4.
- Add [5; −4] to A to get B.
Answer →AB = [5; −4].
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
Find →AB for A(1,2), B(5,−1).
Give me a hint
Subtract the starting coordinates.
Compare my reasoning
- 5−1 = 4.
- −1−2 = −3; A+[4; −3] reaches B.
[4; −3].
Look for these in your work
- Use one subtraction order
- Check the endpoint
Find →PQ and →QP for P(−3,−2), Q(2,4).
Give me a hint
Compute end minus start and then reverse.
Compare my reasoning
- →PQ = [5; 6].
- →QP = [−5; −6].
→PQ = [5; 6]; →QP = [−5; −6].
Look for these in your work
- Retain both signs
- Check reversal
A displacement [−4; 7] carries A(6,−3) to B. Find B and verify →AB.
Give me a hint
Add each movement component to its starting coordinate.
Compare my reasoning
- B = (6−4,−3+7) = (2,4).
- B−A = [−4; 7], confirming the given displacement.
B(2,4).
Look for these in your work
- Separate point and vector notation
- Use a reverse check
Common mistakes
- Use terminal minus initial in both components.
- Subtracting a negative coordinate adds its opposite.
What can you explain now?
Which subtraction gives →AB?
Compare with the explanation
Coordinates of B minus coordinates of A.
The displacement must carry the starting point A to the endpoint B.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Which subtraction gives →AB?
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.