Linear law questions are not mainly about drawing a straight line. They test whether a student can transform a relationship into a form that can be read from a straight-line graph, identify what should be plotted, and then recover the original constants from the gradient and intercept.
This guide focuses on the decision-making behind that process. It belongs to the wider Additional Mathematics Hub and works alongside the Geometry and Trigonometry Guide and Coordinate Geometry guide.
1. The target form
The model to aim for is the familiar straight-line equation:
Y = mX + c
Here, X and Y may not be the original x and y. They may be transformed quantities such as x², 1/x, ln x, ln y or another expression. The entire task is to decide what the new X and Y should be.
2. Worked example: y = ax + b
This relationship is already linear. Plot y against x. The gradient is a and the vertical intercept is b.
That sounds trivial, but it establishes the pattern used in every harder example: identify the graph variables, identify the gradient, identify the intercept, and then connect those quantities back to the constants in the original relationship.
3. Worked example: y = ax² + b
The graph of y against x is not generally a straight line. But if we define X = x², then:
y = aX + b.
Therefore plot y against x². The gradient is a and the vertical intercept is b.
The important move was recognising that x² should become the horizontal variable. A student who simply plots y against x has not transformed the relationship into linear form.
4. Worked example: y = a/x + b
Rewrite the equation as:
y = a(1/x) + b.
Let X = 1/x. Then y = aX + b.
So plot y against 1/x. The gradient is a and the vertical intercept is b.
5. Worked example: y = ax^n
This form is not linear in x and y. Taking logarithms gives:
ln y = ln a + n ln x.
Rearrange mentally into straight-line form:
ln y = n(ln x) + ln a.
Therefore plot ln y against ln x. The gradient is n and the vertical intercept is ln a. If the intercept is known, recover a by exponentiating: a = e^(intercept).
6. Worked example: y = ab^x
Take natural logarithms:
ln y = ln a + x ln b.
So plot ln y against x. The gradient is ln b and the vertical intercept is ln a.
Hence:
- a = e^(intercept)
- b = e^(gradient)
The graph does not directly give a and b. It gives transformed quantities. That distinction is a common source of lost marks.
7. How to decide what to plot
- Start from the original relationship.
- Manipulate it until one expression can be written as a constant times another expression plus a constant.
- Name those expressions mentally as Y and X.
- Write the transformed equation in Y = mX + c form.
- State exactly what is plotted on each axis.
- State what the gradient and intercept represent.
- Convert those transformed quantities back to the original constants.
8. Reading a graph safely
Suppose a graph of ln y against ln x has gradient 2.4 and vertical intercept 0.7. For a model y = ax^n:
- n = 2.4
- ln a = 0.7
- a = e^0.7
Do not report a = 0.7. The intercept is ln a, not a.
9. Gradient discipline
When a straight-line graph is supplied, the gradient should be found from two suitably separated points on the best-fit line, not from two raw data points unless the question specifically instructs otherwise. Write the coordinate differences clearly and keep track of the actual transformed axis values.
If the horizontal axis is 1/x or ln x, use those plotted values in the gradient calculation. Do not silently switch back to x.
10. Common mistakes
| Mistake | Why it happens | Repair |
|---|---|---|
| Plotting y against x automatically | The student sees a graph question and skips transformation | Write Y = mX + c first |
| Using the intercept as the original constant | The transformed meaning is forgotten | Write c = ln a or c = function(a) explicitly |
| Using the wrong horizontal values for gradient | Switching back to original x | Use the coordinates actually shown on the transformed graph |
| Taking logs inconsistently | Only part of a product or power is transformed | Apply logarithm laws line by line |
| Finding gradient from points too close together | Convenience overrides accuracy | Choose separated points on the best-fit line |
11. Practice set
- For y = ax³ + b, state what should be plotted to obtain a straight line, and state the gradient and intercept.
- For y = a/x² + b, state the transformed horizontal variable.
- For y = ax^n, a graph of ln y against ln x has gradient 1.8 and intercept 1.2. State n and express a exactly in exponential form.
- For y = ab^x, a graph of ln y against x has gradient 0.4 and intercept -0.2. Express a and b in exponential form.
- Explain why a straight line after transformation is useful even when the original relationship is nonlinear.
12. Answers
- Plot y against x³; gradient a; vertical intercept b.
- Use 1/x² as the horizontal variable.
- n = 1.8 and a = e^1.2.
- a = e^-0.2 and b = e^0.4.
- The transformation converts a nonlinear relationship into a form where gradient and intercept can reveal parameters systematically.
13. What mastery looks like
A student has mastered linear law when they can start from an unfamiliar relationship, create a valid straight-line form, state the axes without guessing, interpret the gradient and intercept correctly, and recover the original constants. They should also be able to explain why their transformation works.
Continue through Coordinate Geometry in Additional Mathematics, the Geometry and Trigonometry Guide, or return to the A-Math Hub.