AQA GCSE Mathematics Paper 2 (Higher), June 2025: Question 23

5 marks · Hard difficulty · Multi-step Problem

Given quadrilateral ABCD with intersecting diagonals at P, vector AB = 9b - 2a, vector BC = 5a, vector DC = 10b, and ratio BP : PD = 3 : 2, find the value of k if AP : PC = 1 : k.

Practise this question

Question

A geometric diagram of quadrilateral ABCD with diagonals AC and BD intersecting at point P. Directed line segments show vector AB = 9b - 2a, vector BC = 5a, and vector DC = 10b. The text states that the ratio BP to PD is 3 to 2, and the ratio AP to PC is 1 to k. The task asks to work out the value of k, showing all working.

Mark scheme

Show the mark scheme Mark scheme for question 23 worth 5 marks: B1 for vector AC = 3a + 9b, BD = 5a - 10b, or AD = 3a - b; B2 for BP = 3a - 6b or PD = 2a - 4b; B3 for AP = a + 3b or PC = 2a + 6b; B4 for two of AP = a + 3b, PC = 2a + 6b, or AC = 3a + 9b; B5 for final answer k = 2.

How to answer it

Vector Proofs & Ratio Division in Quadrilaterals

📋 What this question tests

This question assesses advanced vector geometry skills on the Higher tier GCSE paper:

  • Finding paths along a geometric diagram using vector addition and subtraction.
  • Dividing a vector into given ratio parts ( BP : PD = 3 : 2 means BP = 3/5 BD ).
  • Forming alternative route expressions for sub-segments (e.g. AP = AB + BP ).
  • Comparing collinear vector components to deduce the ratio scalar k .

Question 23 (5 Marks)

Finding the ratio parameter k along diagonal AC

📐 Step-by-Step Calculation

  1. Find the full vector AC:
    Follow the path from A to C via B:
    AC = AB + BC = (9b - 2a) + 5a = 3a + 9b
  2. Find the full diagonal vector BD:
    Follow the path from B to D via C:
    BD = BC + CD
    Since DC = 10b , going from C to D is -DC = -10b .
    BD = 5a - 10b
  3. Find the segment BP using the given ratio:
    Given BP : PD = 3 : 2 , the line BD has 3 + 2 = 5 parts in total.
    Therefore, BP = (3/5) BD = (3/5)(5a - 10b) = 3a - 6b .
  4. Find vector AP:
    Take the route from A to P via B:
    AP = AB + BP = (9b - 2a) + (3a - 6b)
    Simplifying: AP = a + 3b
  5. Find vector PC:
    Route 1: PC = AC - AP = (3a + 9b) - (a + 3b) = 2a + 6b
    (Alternatively, use PC = PD + DC = (2/5 BD) + 10b = 2a - 4b + 10b = 2a + 6b )
  6. Determine the value of k:
    We are given that AP : PC = 1 : k .
    Notice: PC = 2(a + 3b) = 2 AP .
    Since PC is twice as long as AP , the ratio is 1 : 2 .
    Therefore, k = 2 .

✅ Final Answer

k = 2

Key vectors required for full marks:

  • AP = a + 3b
  • PC = 2a + 6b (or AC = 3a + 9b )

💡 Mark Scheme Breakdown

  • B1: Finds full vector AC = 3a + 9b or BD = 5a - 10b (or AD = 3a - b ).
  • B2: Correct sub-segment on BD: BP = 3a - 6b or PD = 2a - 4b .
  • B3: Correct vector expression for either AP = a + 3b or PC = 2a + 6b .
  • B4: Two correct vectors out of AP , PC , or AC .
  • B5: Fully correct final value: k = 2 with supported working.

🧠 Exam Technique

  • Fraction of a vector: Always add ratio parts together to form the denominator. A ratio of 3 : 2 gives fractions 3/5 and 2/5 , not 3/2 .
  • Check direction arrows: Vector DC = 10b points towards C. Going from C to D means reversing the arrow, giving -10b .
  • State the path clearly: Writing down your route (e.g. AP = AB + BP ) prevents sign mistakes and secures method credit even if arithmetic errors happen later.

❌ Common Errors

  • Inverting ratio: Finding k = 1/2 by computing AP/PC instead of PC/AP . The question asks for AP : PC = 1 : k , meaning PC = k × AP . This error is capped at 4 marks.
  • Sign slip on DC: Adding +10b when going from C to D instead of subtracting 10b .
  • Unsimplified expressions: Leaving vectors as unsimplified expressions like 9b - 2a + 5a , which do not earn vector marks on their own without simplification.
Examiner Insight: High-scoring students systematically solved for the cross-diagonal BD first. Notice that AP = a + 3b and AC = 3(a + 3b) , which immediately confirms that P divides AC in the ratio 1 part to 2 parts (since 1 + 2 = 3 parts in total).

Topics

Geometry and measures · Ratio, proportion and rates of change · 3.4.3 Vectors · 3.3 Ratio, proportion and rates of change

Question and mark scheme from the AQA GCSE Mathematics examination, Paper 2 (Higher), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.