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Parallel and perpendicular

Free Printable Parallel and perpendicular PDF Worksheets, Revision Guide & GCSE Past Paper Questions (Grade 6)

Download free parallel and perpendicular worksheets (PDF), GCSE maths past paper questions and worked solutions for Parallel and perpendicular (6F). This is a Grade 6 algebra topic from the GCSE questions by topic collection for Higher tier GCSE Maths (Edexcel, AQA, OCR and IGCSE).

The free PDF worksheets table below brings together printable question booklets (Qs) and worked answers or mark schemes (Ans) from IGCSE MT, Maths Genie, Corbett Maths, Maths Made Easy, 1st Class Maths, Metatutor, Maths Takeaway, Dr Austin, Naiker Maths, E Classroom and PMT, along with past paper question booklets.

Below the worksheet links you can read our parallel and perpendicular revision notes and worked examples with diagrams, try real GCSE past paper parallel and perpendicular questions with step-by-step solutions, or download a PDF of all parallel and perpendicular worksheet links.

Free Parallel and perpendicular PDF Worksheets & Answers

Grade 6 • Topic 6F
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Grade 6 topics
6A - Recurring decimals6B - Harder indices
6C - The product rule6D - Repeated change
6E - Triple brackets6F - Parallel and perpendicular
6G - Inequalities on graphs6H - Similar shapes (area and vol)
6J - Negative scale factors6K - Circle theorems
6L - Cumulative frequency6M - Box plots
6N - Combined transformations6P - Capture recapture
GCSE Maths Grade 6 Revision Notes & Method

Guide to Parallel and perpendicular: Explanation, Diagrams & Worked Examples

Download Printable One Page Guide (PDF)

Explanation:

Always rearrange any straight-line equation into the form y = mx + c first so you can read its gradient m. Parallel lines have the exact same gradient (m1 = m2).

Perpendicular lines meet at 90° and their gradients are negative reciprocals of each other: m2 = −1m1 (flip the fraction upside down and change the sign, so m1 × m2 = −1).

Example 1: Find the equation of the line parallel to 2y − 6x = 5 that passes through (1, −2).

Solution:

Rearrange: 2y = 6x + 5 → y = 3x + 2.5, so m = 3
Substitute m = 3 and (1, −2) into y = mx + c:
−2 = 3(1) + c → c = −5
Answer: y = 3x − 5

Example 2: Find the equation of the line perpendicular to y = 23x + 4 passing through (4, −1).

m = 2/3 m = −3/2 (4, −1)

Solution:

Perpendicular gradient m⊥ = −32 (negative reciprocal)
Sub m = −32 and (4, −1): −1 = −32(4) + c → c = 5
Answer: y = −32x + 5
Scanned GCSE Exam Questions & Worked Solutions

GCSE Past Paper Parallel and perpendicular Questions & Answers

Past Paper Example 1 GCSE Jun 2025 2H Q20 5 marks
Full Paper (PDF) • Mark Scheme (PDF)
In the diagram
A is the point (0, 8)
B is the point (16, 0)
The point D divides the line segment AB in the ratio 1 : 3
The line **L** passes through D.
The gradient of **L** is √(3)
**L** passes through the point with coordinates (-2, f)
xyODLAB
Show that f < -4
Worked Solution & Mark Scheme (5 marks)
D divides AB in ratio 1 : 3 ⇒ D = (0 + 1/4 × 16, 8 + 1/4 × (0 – 8)) = (4, 6) M2
Equation of L: y – 6 = √3(x – 4) M1
Substitute (–2, f): f – 6 = √3(–2 – 4) = –6√3 ⇒ f = 6 – 6√3 ≈ 6 – 10.392 = –4.392 M1
Since –4.39 < –4, shown. C1
Answer: Shown (f = 6 – 6√3 ≈ –4.39 < –4)
Past Paper Example 2 AQA Nov 2019 1H Q22 3 marks
Full Paper (PDF) • Mark Scheme (PDF)
(-3, 10) is a point on line L.
(4, 0) and (6, 10) are points on line M.
L and M are parallel.
Not drawnaccuratelyxyOM(4, 0)(6, 10)L(-3, 10)
Work out the equation of line L.
Give your answer in the form y = mx + c
Worked Solution & Mark Scheme (3 marks)
Gradient of M = 10 - 06 - 4 = 102 = 5 M1
Since L is parallel to M, gradient of L is also 5.
Equation of L using (-3, 10): M1
y - 10 = 5(x - (-3)) = 5(x + 3) = 5x + 15
y = 5x + 25. A1
Answer: Answer y = 5x + 25
Past Paper Example 3 IGCSE Jan 2020 2H Q19 5 marks
Full Paper (PDF) • Mark Scheme (PDF)
(a) Equation of line parallel to y = 7 − 4x (1) (b) Line perpendicular to L through (−3, 1) and (2, −2) passing through (−6, 4) in form ax + by + c = 0 (4)
(a) Write down an equation of a line that is parallel to the line with equation y = 7 - 4x [1 mark]
(b) Find an equation of the line that is perpendicular to L and passes through the point with coordinates (-6, 4)
Give your answer in the form ax + by + c = 0 where a, b and c are integers. [4 marks]
Worked Solution & Mark Scheme (5 marks)
Part (a) Solution:
Any line with gradient −4, e.g. y = −4x B1
Answer (a): y = y = -4x
Part (b) Solution:
mL = (−2 − 1)/(2 − (−3)) = −3/5 ⇒ mperp = 5/3 M3
y − 4 = (5/3)(x + 6) ⇒ 3y − 12 = 5x + 30 ⇒ 5x − 3y + 42 = 0 A1
Answer (b): 5x - 3y + 42 = 0

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