Linear Models (AI SL)
Linear models describe a constant rate of change - a straight-line relationship between an input and an output, like distance and taxi fare, or hours worked and pay. This topic covers finding the equation of a line from given information, interpreting the gradient and intercept in context, using linear models to solve break-even and comparison problems, and solving systems of linear equations when two conditions must hold at once.
What the syllabus says
This topic maps onto two points in the official IB Applications & Interpretation syllabus, both within the Functions & Modelling unit.
| Code | Syllabus content |
|---|---|
| SL2.1 | Different forms of the equation of a straight line: \(y=mx+c\) (gradient-intercept form), \(ax+by+d=0\) (general form), \(y-y_1=m(x-x_1)\) (point-gradient form). Gradient and intercepts. Parallel lines have \(m_1=m_2\); perpendicular lines have \(m_1\times m_2=-1\). |
| SL2.5 | Modelling with linear functions \(f(x)=mx+c\), including piecewise linear models, for example horizontal distances of an object to a wall, depth of a swimming pool, or mobile phone charges. |
SL2.1 is shared content with Analysis & Approaches; SL2.5 (linear modelling) is specific to Applications & Interpretation.
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is the gradient of a linear model?
The gradient is the constant rate of change of a linear model - how much the output changes for every one-unit increase in the input. In context, it's the "per unit" price, speed, or rate.
e.g. For \(C(n)=0.30n+15\), the gradient \(0.30\) means each extra copy costs \$0.30 more.
What does the \(y\)-intercept represent in a linear model?
The \(y\)-intercept is the value of the output when the input is zero - in a real context, this is often a fixed starting value, such as an initial amount or a base fee, before anything variable is added.
e.g. For \(D(t)=50-2.5t\), the \(D\)-intercept \(50\) cm is the initial depth of water.
How do you find the equation of a line from two points?
Find the gradient using \(m=\dfrac{y_2-y_1}{x_2-x_1}\), then substitute one of the points into \(y=mx+c\) and solve for \(c\).
e.g. \(f(0)=3,\ f(5)=13\): \(m=\dfrac{13-3}{5-0}=2\), so \(f(x)=2x+3.\)
What is a break-even point?
A break-even point is where a cost model and a revenue model give the same value - profit is exactly zero there. Find it by setting \(C(n)=R(n)\) and solving for \(n\).
e.g. \(C(n)=3n+500,\ R(n)=8n\): \(8n=3n+500\Rightarrow5n=500\Rightarrow n=100.\)
When are two lines parallel or perpendicular?
Two lines are parallel when they share the same gradient (\(m_1=m_2\)), and perpendicular when their gradients multiply to \(-1\) (\(m_1\times m_2=-1\)).
e.g. Gradients \(20\) and \(-0.05\): \(20\times(-0.05)=-1\), so the lines are perpendicular.
Key formulas
Four relationships cover almost every question on this topic. The tables below summarise them at a glance - the explanations underneath go into more depth on each one.
Formula reference
These are treated as prior knowledge from earlier study of straight lines rather than formulas printed in the AI booklet, so it's worth having them ready without looking them up.
| Formula | Used for | Booklet? |
|---|---|---|
| \(y=mx+c\) | Gradient-intercept form of a line | Not in booklet - prior knowledge |
| \(m=\dfrac{y_2-y_1}{x_2-x_1}\) | Gradient from two points | Not in booklet - prior knowledge |
| \(m_1=m_2\) | Condition for parallel lines | Not in booklet - prior knowledge |
| \(m_1\times m_2=-1\) | Condition for perpendicular lines | Not in booklet - prior knowledge |
Parallel vs perpendicular
Both conditions compare two gradients - the difference is in exactly how they compare.
| Feature | Parallel lines | Perpendicular lines |
|---|---|---|
| Gradient condition | \(m_1=m_2\) | \(m_1\times m_2=-1\) |
| Visually | Never meet - same steepness | Meet at a right angle |
| Example | \(m=3\) and \(m=3\) | \(m=20\) and \(m=-0.05\) |
The equation of a straight line
Every linear model starts from finding, or being given, the equation of a line.
Gradient-intercept form
\[y=mx+c\]
\(m\) is the gradient, \(c\) is the \(y\)-intercept - the most useful form for reading off both at a glance.
Finding the equation from two points
Calculate the gradient from the two points, then substitute either point into \(y=mx+c\) to solve for \(c\).
Parallel and perpendicular lines
Compare gradients: equal gradients mean parallel lines; gradients that multiply to \(-1\) mean perpendicular lines.
Linear models in context
Once a straight line represents a real situation, its gradient and intercept both need a sentence of interpretation, not just a number.
Interpreting gradient and intercept
The gradient is a rate ("per unit" of input); the intercept is the value at zero input, often a fixed cost or starting amount.
Restricting the domain
A real model is only valid over a sensible domain - state it explicitly, and don't extrapolate the line far beyond it without checking the context still makes sense.
Break-even and comparing two models
Set two linear models equal to find where they agree (break-even, or equal cost); evaluate both at a specific input to see which is smaller beyond that point.
Systems of linear equations
When two unknowns must satisfy two conditions simultaneously, you need a system of equations rather than a single line.
Setting up two equations
Translate each piece of given information into an equation in the same two variables.
Solving by substitution
Rearrange one equation for one variable, substitute it into the other, and solve - then back-substitute to find the second variable.
Solving with the GDC
Enter the coefficients into the calculator's simultaneous equation solver to get both values directly, without algebra by hand.
Worked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
A printing firm charges a fixed set-up fee plus a cost per copy. 100 copies cost \$45 and 250 copies cost \$90.
(a) Find the cost per copy.
(b) Find the set-up fee.
(c) Write the cost model \(C(n)\) for \(n\) copies.
Worked solution
(a) Cost per copy (gradient). The cost rises in a straight line, so \(m = \frac{90-45}{250-100} = \frac{45}{150}\) M1
\(= $0.30 \text{ per copy}.\) A1
(b) Set-up fee (intercept). Substitute a known point into \(C = 0.30n + f\): \(45 = 0.30(100) + f \Rightarrow f = 45 - 30\) M1
\(= $15.\) A1
(c) Model. \(C(n) = 0.30n + 15.\)A1
Service P charges \$6 plus \$1.20 per km. Service Q charges \$10 plus \$0.80 per km.
(a) Write both cost models.
(b) Find the distance at which the costs are equal.
(c) Which is cheaper for a 15 km trip?
Worked solution
(a) Models. \(P(d) = 1.2d + 6\) and \(Q(d)\) A1
\(= 0.8d + 10.\) A1
(b) Equal cost. \(1.2d + 6 = 0.8d + 10 \Rightarrow 0.4d = 4 \Rightarrow d\) M1
\(= 10 \text{ km}.\) A1
(c) At 15 km. \(P(15) = 1.2(15)+6 = 24,\ Q(15) = 0.8(15)+10\) A1
\(= 22.\) So Q is cheaper. R1 Q’s lower per-km rate wins for longer trips.
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Swapping the numerator and denominator when finding the gradient. \(m=\dfrac{y_2-y_1}{x_2-x_1}\), not \(\dfrac{x_2-x_1}{y_2-y_1}\) - keep the same order (point 2 minus point 1) on both top and bottom.
- Checking \(m_1=m_2\) for a perpendicular condition, or the reverse. Parallel needs equal gradients; perpendicular needs gradients that multiply to \(-1\) - mixing the two conditions up is a common slip under time pressure.
- Forgetting to interpret the gradient and intercept, not just calculate them. A question asking to "interpret" or "explain" the gradient wants a sentence linking it back to the real-world rate, not just the number \(m\) restated.
- Rounding a break-even quantity to the nearest whole number without thinking about direction. If \(n=34.8\) chairs are needed to break even, you need \(35\) whole chairs to make a profit, not \(34\) (rounding down under-shoots the target).
Using your GDC
Every step below is a real button sequence, not a vague "use your calculator" hint - covering the TI-84 Plus, TI-Nspire, and Casio fx-9860/fx-CG50. Pick your model to filter down to just the steps that apply to you.
Solve two or three simultaneous equations (linear systems) without elimination by hand.
- Write each equation in the form \(ax+by(+cz)=d\).
- APPS → PlySmlt2 → Simultaneous Eqn Solver; set the number of equations/unknowns and enter the coefficients.TI-84
- menu → Algebra → Solve System of Equations, or use linSolve.Nspire
- Main menu → Equation → Simultaneous, set the number of unknowns, enter the coefficients, SOLVE.Casio
Tip: No solution or infinitely many? The calculator will flag it - that means the lines are parallel or coincident.
See the full GDC guide for more calculator models and topics.
Ready to practise properly?
Linear models questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
How do I find the equation of a linear model from two data points?
Find the gradient using \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\), then substitute either point into \(y = mx + c\) and solve for \(c\).
What does the gradient of a linear model represent?
The gradient is the rate of change of the output per unit of input - a cost per item, a speed, a rate of drainage. Its units come from dividing the output's units by the input's units.
How do I find a break-even point?
Set the cost and revenue functions equal to each other, \(C(n) = R(n)\), and solve for \(n\). That's the point where profit is exactly zero.
Can my GDC solve simultaneous linear equations for me?
Yes - every AI SL calculator has a simultaneous equation or system solver. Enter the coefficients from each equation and it returns the solution directly, without needing substitution or elimination by hand.
Sub-topics
Linear Models broken down into its individual skills, each with its own focused page.
Related topics
More Functions & Modelling topics from the same AI SL syllabus unit, in case you want to keep going.