Step-by-step worked examples and graded practice questions on gradient — calculating it from two points, reading it from a graph, and interpreting it as a rate of change.
📚 Foundation & Higher✅ 15 Practice Questions🔍 Full Worked Examples⚠️ Common Mistakes
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The gradient of a line measures how steep it is — how much y changes for every unit change in x. A positive gradient slopes upward from left to right; a negative gradient slopes downward. The bigger the size of the gradient (ignoring its sign), the steeper the line.
At GCSE you need to be able to:
Calculate the gradient between two coordinates
Read the gradient directly from a graph, using rise over run
Recognise the gradients of horizontal and vertical lines
Interpret gradient as a rate of change in a real-life context
Calculating gradient from two points
Given two coordinates (x₁, y₁) and (x₂, y₂), the gradient is:
m = (y₂ − y₁) / (x₂ − x₁)
This is often remembered as "change in y over change in x", or "rise over run".
Worked Example 1 — Gradient between two points
Find the gradient of the line passing through (2, 5) and (6, 13)
1
Label the points: (x₁, y₁) = (2, 5) and (x₂, y₂) = (6, 13)
2
Substitute: m = (13 − 5) / (6 − 2) = 8 / 4 = 2
AnswerGradient = 2
The line through (2, 5) and (6, 13): rise = 8, run = 4, so gradient = 8 ÷ 4 = 2.
Reading gradient from a graph
To find the gradient directly from a graph, pick two clear points the line passes through — ideally where it crosses grid lines. Count how far you move up (the rise) and how far you move across (the run) between the two points, then divide: gradient = rise ÷ run. Moving down counts as a negative rise; moving left counts as a negative run.
Worked Example 2 — Rise over run
A straight line on a graph rises 6 units for every 3 units it moves to the right. Find its gradient.
1
Rise = 6, run = 3
2
Gradient = rise ÷ run = 6 ÷ 3 = 2
AnswerGradient = 2
Rise 6, run 3: gradient = 6 ÷ 3 = 2.
Horizontal and vertical lines
Two special cases come up often at GCSE:
A horizontal line (e.g. y = 4) has gradient 0 — it does not rise or fall at all.
A vertical line (e.g. x = 7) has an undefined gradient — the run is 0, and dividing by 0 is not possible.
Worked Example 3 — Special lines
State the gradient of the line y = 4, and the gradient of the line x = 7
1
y = 4 is a horizontal line — every point has the same y-value, so it never rises. Gradient = 0
2
x = 7 is a vertical line — every point has the same x-value, so the run is 0. Gradient = undefined
Answery = 4: gradient 0. x = 7: gradient undefined.
The horizontal line y = 4 has gradient 0; the vertical line x = 7 has an undefined gradient.
Gradient as a rate of change
In real-life graphs, the gradient represents a rate of change — for example, speed on a distance-time graph, or cost per mile on a taxi-fare graph. The units of the gradient come from dividing the units on the y-axis by the units on the x-axis.
Worked Example 4 — Gradient in context
A tank fills at a constant rate. The graph of volume (litres) against time (minutes) passes through (0, 0) and (5, 20). Find the rate at which the tank fills.
1
This is a gradient calculation: m = (20 − 0) / (5 − 0) = 20 / 5 = 4
2
The units are litres ÷ minutes, so the tank fills at a rate of 4 litres per minute
Answer4 litres per minute
Volume against time for the filling tank: the gradient (4 litres/min) is the fill rate.
Practice questions
Work through each question before checking the answers. Difficulty is shown for each question.
Foundation (Grade 3–5)
Q1Find the gradient of the line passing through (1, 2) and (4, 11)Foundation
Q2Find the gradient of the line passing through (0, 5) and (5, 0)Foundation
Q3State the gradient of the line y = 6Foundation
Q4State the gradient of the line x = −2Foundation
Q5A line rises 8 units for every 4 units it runs. State its gradient.Foundation
Higher (Grade 5–7)
Q6Find the gradient of the line passing through (−3, 4) and (1, −8)Higher
Q7Find the gradient of the line passing through (2, 7) and (2, 15)Higher
Q8A taxi fare graph (cost in £ against miles) passes through (0, 3) and (10, 23). Find the gradient and interpret it.Higher
Q9Find the gradient of the line joining (−5, −2) and (3, 6)Higher
Q10A temperature graph (°C against hours) passes through (0, 20) and (4, 12). Find the gradient and interpret it.Higher
Higher — Hard (Grade 8–9)
Q11The line through (k, 3) and (5, 11) has gradient 4. Find k.Grade 8–9
Q12A line passes through (2, −1) and has gradient −3/2. Find the y-coordinate when x = 6.Grade 8–9
Q13Points A(1, 2), B(4, 11) and C(7, 20) — show that they are collinear by comparing gradients.Grade 8–9
Q14A cyclist's distance-time graph (km against hours) passes through (0, 0) and (3, 21). Find the gradient and state what it represents.Grade 8–9
Q15The line through (−2, k) and (4, 10) has gradient 5/3. Find k.Grade 8–9
Answers
Foundation (Q1–Q5)
Q13((11 − 2) / (4 − 1) = 9/3)
Q2−1((0 − 5) / (5 − 0) = −5/5)
Q30(horizontal line)
Q4Undefined(vertical line)
Q52(8 ÷ 4)
Higher (Q6–Q10)
Q6−3((−8 − 4) / (1 − (−3)) = −12/4)
Q7Undefined(both points have x = 2, so this is a vertical line)
Q8Gradient = 2; the taxi charges £2 per mile((23 − 3) / (10 − 0) = 20/10)
Q91((6 − (−2)) / (3 − (−5)) = 8/8)
Q10Gradient = −2; the temperature falls by 2°C per hour((12 − 20) / (4 − 0) = −8/4)
Higher — Hard (Q11–Q15)
Q11k = 3(8/(5−k) = 4 → 5−k = 2 → k = 3)
Q12y = −7((y+1)/4 = −3/2 → y+1 = −6 → y = −7)
Q13Gradient AB = gradient BC = 3, so A, B and C are collinear(AB: (11−2)/(4−1)=3; BC: (20−11)/(7−4)=3)
Q14Gradient = 7; this represents the cyclist's speed of 7 km/h((21 − 0) / (3 − 0) = 7)
These are the errors Alamin sees most frequently with gradient at GCSE. Recognising them now will save you marks in the exam.
Common Mistake 1
Flipping the gradient fraction
The formula is m = (y₂ − y₁) / (x₂ − x₁) — change in y over change in x. Writing it upside down gives the reciprocal of the correct gradient, not the correct answer.
Common Mistake 2
Sign errors with negative coordinates
When subtracting negative coordinates, always use brackets: (−8) − (−3) = −8 + 3 = −5, not −11. Rushing this step is the single biggest source of gradient errors.
Common Mistake 3
Mixing up horizontal and vertical line gradients
A horizontal line (y = a number) has gradient 0. A vertical line (x = a number) has an undefined gradient. Students frequently swap these two round.
Common Mistake 4
Not simplifying the gradient fraction
A gradient of 6/4 should always be simplified to 3/2. Leaving fractions unsimplified can cost marks and makes later steps (like finding a perpendicular gradient) harder.
Common Mistake 5
Confusing steepness with the sign of the gradient
A gradient of −5 is steeper than a gradient of 1, even though −5 is a smaller number. Steepness depends on the size of the gradient, ignoring its sign — the sign only tells you the direction of slope.
Common Mistake 6
Losing the units in a rate-of-change question
When a gradient represents a real-life rate (like £ per mile or litres per minute), always state the units in your answer — a bare number is not a full interpretation and can lose marks.
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