AQA AS Level Mathematics Paper 2, June 2025: Question 4

Mark scheme

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Q Marking instructions AO Marks Typical solution

4(a) 2 1.1b B1 2

States − Gradient = −

Subtotal 1

4(b)(i) Obtains perpendicular gradient 1.1b B1 ∴ 3

Gradient of AB =

of

2 p − −(2) 3

PI =

Or 10 − 2 2

Obtains midpoint of AB as p =10

p −2

6 ,

Forms an equation in p by using

coordinates of A and B and their 3.1a M1

gradient of AB (≠ their gradient

from (a))

Or

Substitutes their midpoint into

the equation of L

PI by correct value of p

Or

Forms an equation for AB and

substitutes (10 , p)

Obtains p = 10 1.1b A1

Subtotal 3

4(b)(ii) Writes their correct midpoint 1.1b B1F (6,4)

FT their value of p

Subtotal 1

Question 4 Total 5

How to answer it

Coordinate Geometry: Perpendicular Bisectors

📌 What this question tests

This question assesses core AS-level coordinate geometry skills from Pure Mathematics:

Rearranging linear equations of the form ax + by = c into gradient-intercept form ( y = mx + c ).
Applying the perpendicular gradient rule: m₁ × m₂ = -1 (negative reciprocal).
Understanding the geometric definition of a perpendicular bisector (it passes through the midpoint and meets the segment at 90°).
Finding midpoints and setting up linear algebraic equations to solve for unknown coordinates.

Question 4 (a)

Find the gradient of line L: 2x + 3y = 24 [1 mark]

📐 Step-by-Step Calculation

Rearrange the equation to make y the subject ( y = mx + c ):

1 Subtract 2x from both sides:
3y = -2x + 24

2 Divide every term by 3 :
y = -(2/3)x + 8

3 Read off the coefficient of x :
Gradient = -2/3 (or -0.667)

✅ Mark Scheme Answer

Gradient = -2/3

[B1] (AO 1.1b) Correct value stated in any equivalent exact form (e.g. -2/3 or -4/6).

❌ Common Errors

Sign Error: Forgetting the negative sign when moving 2x over the equals sign, giving 2/3 .
Partial Division: Dividing only the constant by 3 and leaving -2x intact.
Reciprocal Mix-up: Writing -3/2 directly without properly rearranging.

Question 4 (b)(i)

Point A(2, -2) and Point B(10, p). L is the perpendicular bisector of AB. Find the value of p [3 marks]

💡 Key Knowledge

A perpendicular bisector has two defining properties:

Perpendicular: Its gradient is the negative reciprocal of line AB:
m_AB = -1 / m_L = -1 / (-2/3) = 3/2
Bisector: It passes directly through the midpoint of AB:
M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

📐 Method 1: Using the Perpendicular Gradient

1 Find gradient of segment AB using points (2, -2) and (10, p):
Gradient of AB = (y₂ - y₁) / (x₂ - x₁) = (p - (-2)) / (10 - 2) = (p + 2) / 8

2 Equate to the perpendicular gradient ( 3/2 ):
(p + 2) / 8 = 3/2

3 Cross-multiply and solve for p :
2(p + 2) = 24
2p + 4 = 24 ⇒ 2p = 20 ⇒ p = 10

📐 Method 2: Using the Midpoint on Line L

1 Find midpoint of AB in terms of p :
M = ((2 + 10)/2, (-2 + p)/2) = (6, (p - 2)/2)

2 Substitute x = 6 and y = (p - 2)/2 into line L ( 2x + 3y = 24 ):
2(6) + 3((p - 2)/2) = 24
12 + (3p - 6)/2 = 24

3 Solve for p :
(3p - 6)/2 = 12 ⇒ 3p - 6 = 24 ⇒ 3p = 30 ⇒ p = 10

✅ Mark Breakdown

B1: Obtains perpendicular gradient 3/2 (can be implied) OR finds correct midpoint (6, (p - 2)/2) .
M1: Forms an equation in p by equating gradient of AB to 3/2 , or substituting the midpoint into line L.
A1: Correctly computes p = 10.

❌ Common Errors & Pitfalls

Double Negative Slip: Writing p - (-2) as p - 2 instead of p + 2 .
Reusing Parallel Gradient: Setting the gradient of AB equal to -2/3 instead of taking the negative reciprocal 3/2 .
Unfinished Midpoint Method: Finding the midpoint coordinates but failing to substitute them into the line equation.

Question 4 (b)(ii)

Find the coordinates of the point of intersection of AB and L [1 mark]

🧠 Exam Technique Shortcut

You do not need to solve simultaneous equations for lines AB and L!

By definition, a perpendicular bisector intersects the line segment exactly at its midpoint. Simply substitute your value of p = 10 into the midpoint formula.

📐 Step-by-Step Calculation

1 Midpoint formula:
Intersection = ((x_A + x_B)/2, (y_A + y_B)/2)

2 Substitute coordinates A(2, -2) and B(10, 10):
x = (2 + 10) / 2 = 12 / 2 = 6
y = (-2 + 10) / 2 = 8 / 2 = 4

3 Point of intersection is (6, 4).

✅ Mark Scheme Answer

(6, 4)

[B1F] (AO 1.1b) Full follow-through mark allowed: evaluates (6, (p - 2)/2) correctly for their calculated value of p from (b)(i).

Topics

Pure Mathematics · C: Coordinate geometry in the (x, y) plane