Overview
What you’ll learn
- Plot and read points, and read a gradient from a grid as rise over run.
- Compute the gradient between two points as the difference of y-coordinates over the difference of x-coordinates.
- Find midpoints by averaging coordinates, and lengths by Pythagoras’ theorem, leaving exact surd answers where required (Extended).
- Build equations of lines from a gradient and a point, and rearrange ax + by = c to read the gradient (Extended).
- Recognise parallel lines (equal gradients) and construct perpendicular lines and bisectors with gradients multiplying to −1 (Extended).
Syllabus checklist (0580)
Statements are paraphrased and grouped by tier. Core is studied by everyone; Extended content is examined on Papers 2 and 4 only.
Core — every candidate
- 3.1Use Cartesian coordinates in two dimensions to describe and locate points.
- 3.2Draw and interpret straight-line graphs of the form y = mx + c, and read gradients from a grid; rearrange equations like ax + by = c to find the gradient. (Extended for two-point gradients and ax + by = c)
- 3.5Find the equation of a straight line from its gradient and a point on it, in the form y = mx + c.
- 3.6Find the gradient and equation of parallel lines.
Extended — Extended candidates only
- 3.4Calculate the length and the coordinates of the midpoint of a line segment. (Extended)
- 3.7Use perpendicular gradients (products of −1) and find the equation of a perpendicular bisector. (Extended)
In 30 seconds
- 1Gradient = rise ÷ run = (y₂ − y₁) ⁄ (x₂ − x₁): pick two points, subtract consistently, keep the sign.
- 2In y = mx + c, m is the gradient and c is the y-intercept — the line crosses the y-axis at (0, c).
- 3Midpoint (Extended): average the x’s, average the y’s.
- 4Length (Extended): Pythagoras on the horizontal and vertical gaps: √((x₂ − x₁)² + (y₂ − y₁)²).
- 5Parallel lines share a gradient; perpendicular gradients multiply to −1 — flip the fraction and swap the sign (Extended).
- 6A perpendicular bisector (Extended) passes through the midpoint, with the negative-reciprocal gradient of the segment.
- 7To test whether a point lies on a line, substitute its x and check the y comes out.
Notes · 1 of 4
Gradient — steepness with a sign
Rise over run, between any two points you like.
The gradient of a straight line measures how fast it climbs: (change in y) ÷ (change in x). For the line through (1, 2) and (5, 10): (10 − 2) ⁄ (5 − 1) = 8 ⁄ 4 = 2 — up two for every one across. The formula works from either end as long as you subtract in the same order on top and bottom. Downhill lines have negative gradients: (−2, 3) to (4, 0) gives (0 − 3) ⁄ (4 − (−2)) = −3⁄6 = −½.
In the equation form y = mx + c, the gradient m sits in front of the x: y = 5 − 2x has gradient −2 (the minus travels with the 2). Extended lines may arrive dressed as ax + by = c: rearrange to gradient–intercept form first — 3x + 2y = 12 → y = 6 − 1.5x, so the gradient is −3⁄2.
Notes · 2 of 4
Midpoint and length of a segment
Average for the middle, Pythagoras for the stretch. (Extended)
Extended. The midpoint splits a segment into two equal parts: average the x-coordinates, average the y-coordinates. For (2, 3) and (8, 11): ((2 + 8)⁄2, (3 + 11)⁄2) = (5, 7). Averages never mind order, so there is no sign trap here — unlike the gradient, this one is forgiving.
The length of a segment is the hypotenuse of the right triangle its horizontal and vertical gaps make: length = √((x₂ − x₁)² + (y₂ − y₁)²). For (−2, 1) to (4, 9): √(6² + 8²) = √100 = 10. Leave surd answers exact when the question says so — √45 = 3√5 beats a rounded 6.71 — and remember the length is always the positive root.
Notes · 3 of 4
Equations of straight lines
One gradient and one point are enough to write the whole line.
A line is fixed by a gradient and any point on it. For gradient 3 through (2, 1): start from y − y₁ = m(x − x₁), so y − 1 = 3(x − 2), which tidies to y = 3x − 5. Alternatively use y = mx + c and substitute the point to find c: 1 = 3(2) + c → c = −5. Check the point: 3(2) − 5 = 1 ✓.
The intercepts fall out of the equation: the y-intercept is c, and the x-intercept comes from setting y = 0 — for y = 2x − 6, 0 = 2x − 6 gives x = 3. To test whether a point sits on a line, substitute: (4, 11) in y = 3x − 1 gives 3(4) − 1 = 11 ✓ — it lies on the line.
Notes · 4 of 4
Parallel and perpendicular lines
Same gradient copies a line; negative reciprocals cross it at right angles. (Extended for perpendicular)
Parallel lines never meet, so they share a gradient. The line through (1, 4) parallel to 2x + y = 9: rearrange the given line to y = 9 − 2x, gradient −2; the new line is y − 4 = −2(x − 1), i.e. y = −2x + 6, with x-intercept 3. Copy the gradient, then find c from the point.
Extended. Perpendicular gradients multiply to −1: flip the fraction and flip the sign. A gradient of ½ has perpendicular −2; a gradient of −3⁄4 has perpendicular +4⁄3. The perpendicular bisector of a segment combines three skills: find the midpoint, find the segment’s gradient, then write the line through the midpoint with the negative-reciprocal gradient. For P(2, 1) and Q(6, 9): midpoint (4, 5), gradient of PQ is 2, so the bisector has gradient −½ through (4, 5): y − 5 = −½(x − 4), i.e. y = 7 − ½x.
Revise
Key terms
The words examiners expect you to use precisely.
- Gradient
- The steepness of a line: rise divided by run.(1, 2) to (5, 10) → gradient 2.
- y-intercept
- The point where a line crosses the y-axis, the c in y = mx + c.y = 3x − 2 crosses at (0, −2).
- Midpoint (Extended)
- The point halfway along a segment: averages of the coordinates.(2, 3) and (8, 11) → (5, 7).
- Segment length (Extended)
- The distance between two points, by Pythagoras.(−2, 1) to (4, 9) → 10.
- Parallel lines
- Lines with equal gradients that never meet.y = 4x + 1 and y = 4x + 3.
- Perpendicular gradient (Extended)
- The negative reciprocal: m⊥ = −1⁄m.Gradient ½ → perpendicular −2.
Show 2 more key termsShow fewer key terms
- Perpendicular bisector (Extended)
- The line through a segment’s midpoint at right angles to it.Of (0, 2)/(6, 8): y = −x + 8.
- x-intercept
- The point where a line crosses the x-axis (y = 0).y = 2x − 6 crosses at (3, 0).
Revise
Common mistakes
Where students most often lose marks in this topic — and what to do instead.
Mistake: Subtracting coordinates in different orders.
(y₂ − y₁) over (x₁ − x₂) flips the gradient’s sign.
Instead: Match the order on top and bottom: (10 − 2)⁄(5 − 1).
Mistake: Averaging coordinates without dividing by 2.
(2 + 8, 3 + 11) = (10, 14) is not the midpoint.
Instead: Divide each sum by 2: (5, 7).
Show 10 more mistakesShow fewer mistakes
Mistake: Adding the coordinate gaps instead of squaring them.
Distance is not 6 + 8 = 14.
Instead: √(6² + 8²) = √100 = 10.
Mistake: Rounding a surd length when the question says “exact”.
6.71 is not 3√5.
Instead: √45 = √(9 × 5) = 3√5 exactly.
Mistake: Giving parallel lines the same c as well as the same m.
Parallel lines differ in where they cross the axis.
Instead: Copy the gradient only; find c from the new point.
Mistake: Flipping a perpendicular fraction without changing the sign.
The perpendicular to 2⁄5 is not 5⁄2.
Instead: Negative reciprocal: −5⁄2.
Mistake: Building the bisector through one endpoint instead of the midpoint.
The bisector must pass halfway along the segment.
Instead: Midpoint first, then the negative-reciprocal gradient through it.
Mistake: Reading the gradient of ax + by = c as a⁄b.
3x + 2y = 12 has gradient −3⁄2, not 3⁄2.
Instead: Rearrange to y = mx + c before reading m.
Mistake: Asserting a point is on a line without substituting.
Eyeballing a graph is not evidence.
Instead: Substitute the x, compute, compare the y.
Mistake: Adding the coordinate gaps without squaring: 3 + 4 = 7 for a length.
Pythagoras squares each gap before the root.
Instead: √(3² + 4²) = 5, not 7.
Mistake: Reading the y-intercept from where the line crosses the x-axis.
The y-intercept is the point with x = 0, not y = 0.
Instead: y = −2x + 8 meets the y-axis at (0, 8) and the x-axis at (4, 0).
Mistake: Judging collinearity from rough sketching instead of gradients.
A hand-drawn diagram can hide a bend.
Instead: Equal gradients between consecutive points prove collinearity: A to B and B to C both 2.
Check yourself
Quick quiz
24 questions with instant explanations. About 5 minutes.
Question 1 of 3
- Question 1: current question
- Question 2: not answered
- Question 3: not answered
Compute a gradient from two points
The gradient of the line through (1, 2) and (5, 10) is
Ready for more practice on Coordinate geometry?
Generate a fresh 5-question adaptive drill grounded in this topic’s 0580 key points. Free with your study account.
Practise
Topical questions
14 exam-style questions with model answers and mark schemes.
A set of structured and calculation questions matched to this topic, in three tiers, each with a model answer, mark scheme and examiner tip.
Practise topical questionsExam practice
Practice paper
A 30-mark paper set out like a Cambridge paper, with a separate mark scheme. Give yourself 40 min.
Question paper
Topic 11 Practice Paper
30 marks · 40 min · 5 pages
10 multiple-choice questions and structured questions with answer lines. Topic 11: gradients from two points, midpoints and lengths of segments, equations of straight lines and their intercepts, parallel and perpendicular lines, and perpendicular bisectors.
Mark scheme
PROTopic 11 Practice Paper — mark scheme
Answers · Marking points · 3 pages
Model answers with a mark for each creditworthy point — for marking at home or in class.
Which papers examine this topic?
Paper 1 / 3
Core: 80 marks each, 1 h 30 each
Paper 1 non-calculator, Paper 3 calculator
Paper 2 / 4
Extended: 100 marks each, 2 h each
Paper 2 non-calculator, Paper 4 calculator
Accuracy
3 s.f., angles to 1 d.p.
Unless a question states otherwise
Want a full paper? Try the Paper 1-style Core, Paper 2-style non-calculator or Paper 4-style calculator sets.
Questions
Frequently asked questions
How do I remember the gradient formula without mixing up the order?
Write it as (y₂ − y₁)⁄(x₂ − x₁) and label your two points 1 and 2 before subtracting anything. As long as the y-subtraction and x-subtraction use the points in the same order, the sign comes out right. A quick sketch showing whether the line rises or falls confirms the sign.
When should I leave a length as a surd?
Whenever the question says “exact”, “give your answer in surd form”, or the surd simplifies beautifully (√45 = 3√5). If a decimal is wanted, the question states an accuracy such as 3 significant figures — then round once, at the end, from the calculator value.
What is the fastest way to find the perpendicular gradient?
Flip the fraction and change the sign: 2⁄5 becomes −5⁄2, −3⁄4 becomes 4⁄3. Check by multiplying: the original and the perpendicular gradient must give −1.
Do Core candidates need perpendicular bisectors?
No — midpoints, lengths, two-point gradients and perpendicular lines are Extended statements. Core candidates work with gradients read from grids, equations in y = mx + c form, and parallel lines, all of which appear on Core papers.
Is QuizQuip connected to Cambridge?
No. Cambridge IGCSE and Cambridge are trademarks of Cambridge University Press & Assessment. QuizQuip is independent — not affiliated with, endorsed by, or connected to Cambridge. All our notes and questions are original material written against the published 0580 syllabus; we do not host Cambridge past papers.
Up next
Topic 12: Geometry, angles and polygons