A Level Maths: Straight Line Geometry Topic Summary and Resources
Video Lessons
Watch alongside the worksheet for the full lesson experience, then test your understanding with the lesson questions.
Revision Notes
Handwritten notes summarising the key ideas for each lesson. Ideal for quick review before a test.
Exam Questions
Past-paper-style questions organised by topic, with full mark schemes.
Drawn from OCR and Edexcel past papers but designed to be useful for students of all UK exam boards — including AQA and OCR MEI — unless a sheet is explicitly board-specific.
Before You Start This Topic
It will help if you are confident with the following:
- GCSE Maths Coordinate Geometry — plotting points, gradients, and basic line equations from GCSE are assumed
- Quadratic Equations — useful for line-curve intersection problems
A Level Maths straight line geometry introduces coordinate geometry in the $(x, y)$ plane. You learn to work with equations of straight lines fluently in any form, find lines through given points, and use parallel and perpendicular relationships. This is foundational material for Circle Geometry, Differentiation (where tangents and normals are straight lines), and a huge amount of Vectors work.
You handle straight-line equations in the forms $y = mx + c$, $y – y_1 = m(x – x_1)$, and $ax + by + c = 0$, converting between them as needed. Given two points, you find the equation of the line through them; given a point and a gradient, you find the equation. The parallel condition (gradients equal, $m_1 = m_2$) and the perpendicular condition (gradients multiply to $-1$) let you build lines with specific relationships to others. You also model real situations with straight-line equations, including converting between degrees Celsius and degrees Fahrenheit and modelling constant-speed motion.
Straight line geometry is part of the Pure Maths strand of A Level Maths and is fundamental revision content for AQA, Edexcel, OCR, and OCR MEI students.
Watch out for…
A few things to be careful with: the perpendicular condition is $m_1 \times m_2 = -1$, so the perpendicular gradient is the negative reciprocal (the perpendicular to gradient $\tfrac{2}{3}$ has gradient $-\tfrac{3}{2}$, not $-\tfrac{2}{3}$); a vertical line has undefined gradient and cannot be written as $y = mx + c$ (use $x = \text{constant}$ instead); and when finding the equation in $ax + by + c = 0$ form, clear fractions and bring all terms to one side, but check the question for the form it actually wants.