Recognising Linear Relationships
Learn to identify linear relationships from tables, graphs and descriptions. Understand why constant rate of change always produces a straight line.
Printable Worksheets
Print or save as PDF — or build a custom worksheet from any module's questions.
Which of these graphs show a straight line? Which show a constant rate of change?
Reveal answer
Straight lines: A and C. Both have a constant rate of change — they never bend. Graphs B and D are curved, meaning their rate of change varies — they are non-linear.
A linear relationship is one where the dependent variable changes at a constant rate as the independent variable increases.
Three tests for linearity: the graph is a straight line, the first differences in a table are constant, and the gradient is the same no matter which two points you choose.
To test a table of values, check the first differences between consecutive $y$-values. If they are all equal, the relationship is linear.
Watch Me Solve It · 2 examples
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1Calculate first differences$8-5=3$, $11-8=3$, $14-11=3$, $17-14=3$Subtract each $y$-value from the next one.
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2Check if all differences are equalDifferences: 3, 3, 3, 3 — all the sameIf even one is different, the relationship is not linear.
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3ConcludeLinear! Gradient $m = 3$, equation $y = 3x + 2$The constant difference of 3 equals the gradient.
The straight line test is the simplest visual check: if all plotted points form a perfectly straight line, the relationship is linear.
Common mistake: "It looks kind of straight" — use a ruler to check, or verify with the gradient test.
Better approach: If even one point does not fall on the line (and it's not measurement error), the relationship is not linear.
On any straight line, the gradient is the same everywhere. No matter which two points you choose, you always get the same gradient value.
Three points on a line: $A(1,4)$, $B(3,8)$, $C(5,12)$. Gradient $AB = \frac{8-4}{3-1} = 2$. Gradient $BC = \frac{12-8}{5-3} = 2$. Gradient $AC = \frac{12-4}{5-1} = 2$. Always 2.
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1Gradient A to B$$m_{AB} = \frac{11-5}{4-2} = \frac{6}{2} = 3$$
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2Gradient B to C$$m_{BC} = \frac{17-11}{6-4} = \frac{6}{2} = 3$$
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3Gradient A to C (any two points work)$$m_{AC} = \frac{17-5}{6-2} = \frac{12}{4} = 3$$All three pairs give gradient = 3. The relationship is linear!
Comparing $y = 2x + 1$ (linear) with $y = x^2$ (non-linear) side by side:
$y = 2x + 1$ — Linear
| $x$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| $y$ | 1 | 3 | 5 | 7 |
| Diff | — | +2 | +2 | +2 |
Constant differences = Linear
$y = x^2$ — Non-linear
| $x$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| $y$ | 0 | 1 | 4 | 9 |
| Diff | — | +1 | +3 | +5 |
Changing differences = Non-linear
The key question: "Does one quantity change at a constant rate relative to another?"
From a Table
- Calculate first differences ($y_2 - y_1$, etc.)
- If ALL differences are equal: linear
- Common difference = gradient $m$
From a Graph
- Plot all points
- If they lie on a perfectly straight line: linear
- Check with a ruler or gradient calculation
From a Description
- "Per hour", "per kg", "fixed price" suggest linear
- Write as $y = mx + c$
- Constant rate of change = linear
Key Formula
- Gradient $= \dfrac{y_2 - y_1}{x_2 - x_1}$
- Same result for any two points on a straight line
Brain Trainer · 10 problems
Quick-fire questions. Answer each, then reveal.
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1 The first differences of a table are $2, 2, 2, 2$. Is it linear?
Yes — all differences equal 2, so linear with $m = 2$.Yes, linear -
2 First differences: $5, 5, 6, 5$. Is it linear?
No — the 6 breaks the pattern. Not linear.No, not linear -
3 Two points on a line give gradient 4. Two other points give gradient 5. Is it a straight line?
No. A straight line has constant gradient. Different gradients means the graph is curved.No — not linear -
4 Is $y = x^2 + 3$ a linear relationship?
No — $x$ is squared. The graph is a parabola (curve), not a straight line.No, non-linear -
5 A taxi charges \$3 flag fall plus \$2/km. Is this linear? What is the gradient?
Yes, linear. $m = 2$, equation: $C = 2d + 3$.Yes, m = 2 -
6 Points $(2, 3)$ and $(4, 7)$ are on a line. What is the gradient?
$m = \frac{7-3}{4-2} = \frac{4}{2} = 2$2 -
7 Give the first differences for $y$-values: $5, 8, 11, 14, 17$. Linear?
Differences: 3, 3, 3, 3 — all equal, so linear with gradient 3.Yes, linear, m = 3 -
8 Is the area of a square ($A = s^2$) linearly related to its side length $s$?
No — side is squared. Differences are $3, 5, 7$ (not constant).No, non-linear -
9 A phone plan charges \$30/month. Is cost vs months linear? What is $m$?
Yes, linear. $m = 30$, $c = 0$. Equation: $C = 30n$.Yes, m = 30 -
10 Name two different ways to test if a relationship is linear.
(1) First differences test — all consecutive differences equal. (2) Straight line test — plot points and check they form a straight line.First differences; straight line test
Quick Check · 5 questions
| $x$ | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| $y$ | 2 | 5 | 8 | 11 | 14 |
Show Your Working · 3 questions
SAQ 1. Show that the table below represents a linear relationship using the first differences test.
| $x$ | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| $y$ | 4 | 7 | 10 | 13 | 16 |
SAQ 2. Explain in your own words why a straight line has constant gradient. Include a diagram to support your explanation.
SAQ 3. Give a real-world example of a linear relationship and explain in detail why it is linear (not just "it makes a straight line").
Quick Check
1. A — First differences all equal 3. Linear with $m = 3$.
2. B — A linear relationship is defined by a straight-line graph.
3. C — Gradient of a straight line is constant. Always 2, no matter which points.
4. C — \$2/kg is a constant rate of change. Linear.
5. A — Not linear — the difference of 5 breaks the pattern.
Model Answers
SAQ 1: First differences: $7-4=3$, $10-7=3$, $13-10=3$, $16-13=3$. All differences equal 3 → constant → linear. Gradient $m = 3$, equation $y = 3x + 1$.
SAQ 2: A straight line never changes direction. It goes up (or down) at exactly the same angle everywhere. Pick any two points and calculate rise/run — you always get the same answer because the line doesn't bend. Diagram should show two different gradient triangles on the same line with equal ratios.
SAQ 3 (example): A plumber charges \$50 call-out plus \$80/hr. Why linear: cost increases by exactly \$80 for every extra hour — the rate of change is constant. Equation: $C = 80h + 50$. Table would show constant first differences of 80. Other examples: taxi fares with flag fall, unit conversion rules.
Collinearity and Counterexamples
Part A: Three points are $A(1, 4)$, $B(3, 10)$ and $C(5, 16)$. Show these are collinear (on the same straight line) by checking all three gradient pairs.
Part B: A student says checking gradient $AB$ and gradient $BC$ is enough to prove linearity. Can you find a counterexample where this two-pair check misses a non-linear relationship?
Reveal solutions
Part A: $m_{AB} = \frac{10-4}{3-1} = 3$. $m_{BC} = \frac{16-10}{5-3} = 3$. $m_{AC} = \frac{16-4}{5-1} = 3$. All equal → collinear.
Part B: With $(0,0)$, $(1,1)$, $(2,2)$, $(3,5)$: checking only AB and BC gives $m=1$ and $m=1$, but the point $(3,5)$ is off the line through the first three. Always check all pairs or use the full first-differences test.
Linear = straight line
Constant rate of change, constant gradient.
First differences test
All consecutive $y$-differences must be equal.
Constant gradient
Any two points on a straight line give the same gradient.
Non-linear = curved
Changing first differences → curve → not linear.
$y = mx + c$
Linear form. $m$ = gradient. $c$ = $y$-intercept.
Real-world clues
"Per unit", "fixed rate" → linear.
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