What is a linear function in the context of IB MYP Mathematics?

In IB MYP Mathematics, a linear function is an algebraic equation that forms a straight line when graphed on a coordinate plane. It is typically expressed in the form y = mx + b, where m represents the slope of the line and b is the y-intercept. Linear functions are fundamental in understanding relationships between variables and are widely used in various real-world applications.

How do you determine the slope of a linear function?

The slope of a linear function, represented by 'm' in the equation y = mx + b, indicates the steepness and direction of the line. It is calculated as the ratio of the change in the y-coordinate to the change in the x-coordinate between two distinct points on the line. Mathematically, it is expressed as m = (y2 - y1) / (x2 - x1). A positive slope means the line ascends, while a negative slope indicates it descends.

What is the significance of the y-intercept in a linear function?

The y-intercept of a linear function, denoted as 'b' in the equation y = mx + b, is the point where the line crosses the y-axis. It represents the value of y when x is zero. Understanding the y-intercept is crucial as it provides a starting point for graphing the line and helps in interpreting the initial value of a relationship in real-world scenarios.

How can you graph a linear function using its equation?

To graph a linear function from its equation y = mx + b, start by plotting the y-intercept (0, b) on the coordinate plane. Then, use the slope 'm' to determine the next point. For example, if the slope is 2, move up 2 units and 1 unit to the right from the y-intercept. Connect these points with a straight line extending in both directions. This visual representation helps in understanding the behavior of the function.

What are some real-world applications of linear functions?

Linear functions are used extensively in real-world applications such as calculating speed, determining cost, and analyzing trends. For instance, in economics, they model supply and demand relationships. In physics, they describe motion at constant speed. Understanding linear functions allows students to solve practical problems and make predictions based on linear relationships.

Why is "Computing gradient as $\dfrac{\Delta x}{\Delta y}$." a common mistake in Linear Functions?

Why it happens: Mixing up rise and run. How to avoid it: Gradient = RISE/RUN = y_2 - y_1/x_2 - x_1. Y on TOP.

Why is "Writing perpendicular gradient as $\dfrac{1}{m_1}$ (no negative)." a common mistake in Linear Functions?

Why it happens: Missing the sign flip. How to avoid it: Perpendicular = NEGATIVE RECIPROCAL. m_1 · m_2 = -1 requires both: flip AND negate.

Why is "Calling two lines with same $c$ but different $m$ parallel." a common mistake in Linear Functions?

Why it happens: Confusing 'crossing axis at the same place' with 'parallel'. How to avoid it: Parallel requires SAME GRADIENT. y-intercept can be anything.

AlgebraIB MYP Maths Extended Level

Linear Functions

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Linear Functions — IB MYP Mathematics (Extended): straight lines and their equations

A linear function makes a straight-line graph. This MYP Mathematics Extended note covers gradient-intercept form, point-gradient form, finding equations, and parallel / perpendicular relationships.

What you’ll learn

Mapped to the IB MYP Mathematics subject guide (2026 onwards).

MYP Mathematics A — Find gradient from two points.
MYP Mathematics A — Find the equation of a line.
MYP Mathematics A — Identify parallel and perpendicular lines.
MYP Mathematics C — Sketch graphs from the equation.

Gradient-intercept form

$y = mx + c$ — the canonical form.

Gradient-intercept form: $y = mx + c$ .

$m$ = GRADIENT (how steep, and in which direction).
$c$ = Y-INTERCEPT (where the line crosses the y-axis).

The gradient is calculated from any two points $(x_1, y_1)$ and $(x_2, y_2)$ on the line:

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}}.$

Worked example. Find the gradient of the line through $(2, 5)$ and $(8, 17)$ . $m = \frac{17 - 5}{8 - 2} = \frac{12}{6} = 2.$

Interpretation of $m$ :

$m > 0$ : line slopes UP from left to right.
$m < 0$ : line slopes DOWN from left to right.
$m = 0$ : HORIZONTAL line.
Vertical lines have UNDEFINED gradient.

Three line types by gradient sign. Positive: up. Negative: down. Zero: horizontal.

To sketch $y = mx + c$ :

Plot the y-intercept $(0, c)$ .
From there, move $1$ right and $m$ up (or down if $m$ negative).
Draw a line.

Worked example. Sketch $y = -2x + 3$ .

Plot $(0, 3)$ .
Move 1 right, 2 down → $(1, 1)$ .
Draw a line through both points.

$y = mx + c$ — $m$ is gradient, $c$ is y-intercept.
Gradient = rise/run.
Sign of $m$ tells direction.

Finding the equation of a line

Two main forms — pick whichever fits the given info.

Form 1 (gradient + y-intercept): $y = mx + c$ . Directly plug in.

Form 2 (gradient + a point): $y - y_1 = m(x - x_1)$ .

This point-gradient form is incredibly powerful — it works even when you don't know the y-intercept.

Worked example. Find the equation of the line with gradient 3 passing through $(2, 5)$ .

Use point-gradient form: $y - 5 = 3(x - 2)$ .
Rearrange: $y = 3x - 6 + 5 = 3x - 1$ .

Worked example (two points). Find the equation of the line through $(1, 4)$ and $(3, 10)$ .

Step 1 — gradient: $m = \dfrac{10 - 4}{3 - 1} = \dfrac{6}{2} = 3$ .
Step 2 — use one of the points in point-gradient form: $y - 4 = 3(x - 1)$ .
Rearrange: $y = 3x - 3 + 4 = 3x + 1$ .

(Pick the point with simpler numbers if you have a choice.)

Worked example (real context). A taxi charges $5 plus $2 per km. Find an equation for cost $C$ in terms of distance $d$ .

This is $C = 2d + 5$ — linear with $m = 2$ and $c = 5$ .
The y-intercept is the fixed fee; the gradient is the per-km rate.

$y - y_1 = m(x - x_1)$ — most useful form for general lines.
From two points: find gradient first, then use either point in point-gradient form.
Real-context: $c$ = fixed cost, $m$ = rate.

Parallel and perpendicular

Compare gradients to classify line relationships.

Two lines are PARALLEL iff they have the SAME gradient (and different y-intercepts — same gradient and intercept would be the same line).

Two lines are PERPENDICULAR iff their gradients multiply to give $-1$ : $m_1 \times m_2 = -1 \iff m_2 = -\frac{1}{m_1}.$

Equivalently, perpendicular gradients are NEGATIVE RECIPROCALS of each other.

Line 1 gradient	Perpendicular gradient
$2$	$-\dfrac{1}{2}$
$-3$	$\dfrac{1}{3}$
$\dfrac{2}{5}$	$-\dfrac{5}{2}$
$1$	$-1$

Worked example. Find the equation of the line through $(3, 4)$ perpendicular to $y = 2x + 1$ .

Original gradient: $m_1 = 2$ . Perpendicular gradient: $m_2 = -\tfrac{1}{2}$ .
Through $(3, 4)$ : $y - 4 = -\tfrac{1}{2}(x - 3)$ .
$y = -\tfrac{1}{2}x + \tfrac{3}{2} + 4 = -\tfrac{1}{2}x + \tfrac{11}{2}$ .

This is a classic Extended-level question. The key step is recognising that 'perpendicular' translates to the gradient rule.

Parallel: same gradient.
Perpendicular: $m_1 \cdot m_2 = -1$ .
Perpendicular gradient = negative reciprocal of original.

How it’s examined

Criterion A: find equation, gradient, intercept. Criterion C: sketch a line; classify lines as parallel/perpendicular.

Sources: IB MYP Mathematics subject guide (IBO, official). Last reviewed 2026-05-25.

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Step-by-step worked examples — Linear Functions

Step-by-step solutions to past-paper-style questions on linear functions, written exactly the way a tutor would explain them at the board.

1Find gradient from two points
Getting started• gradient
Question
Find the gradient of the line through $(-3, 1)$ and $(2, 11)$ .
Step-by-step solution
1. Step 1
  Use $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ .
2. Step 2
  $m = \dfrac{11 - 1}{2 - (-3)} = \dfrac{10}{5} = 2$ .
Answer
$m = 2$
2Equation from gradient + point
Building confidence• equation
Question
Find the equation of the line with gradient $-2$ passing through $(4, -1)$ .
Step-by-step solution
1. Step 1
  Point-gradient: $y - y_1 = m(x - x_1)$ , so $y - (-1) = -2(x - 4)$ .
2. Step 2
  Simplify: $y + 1 = -2x + 8 \Rightarrow y = -2x + 7$ .
Answer
$y = -2x + 7$
3Equation from two points
Building confidence• equation
Question
Find the equation of the line through $(1, 5)$ and $(-2, -4)$ .
Step-by-step solution
1. Step 1
  Gradient: $m = \dfrac{-4 - 5}{-2 - 1} = \dfrac{-9}{-3} = 3$ .
2. Step 2
  Use point-gradient with $(1, 5)$ : $y - 5 = 3(x - 1)$ .
3. Step 3
  Simplify: $y = 3x - 3 + 5 = 3x + 2$ .
Answer
$y = 3x + 2$
4Perpendicular line
Stretch• perpendicular
Question
Find the equation of the line through $(2, 7)$ perpendicular to $y = \dfrac{1}{3}x + 4$ .
Step-by-step solution
1. Step 1
  Original gradient: $m_1 = \tfrac{1}{3}$ .
2. Step 2
  Perpendicular gradient: $m_2 = -3$ (negative reciprocal).
3. Step 3
  Through $(2, 7)$ : $y - 7 = -3(x - 2)$ .
4. Step 4
  Simplify: $y = -3x + 6 + 7 = -3x + 13$ .
Answer
$y = -3x + 13$

Key Definitions and Keywords — Linear Functions

Definitions to memorise and the exact keywords mark schemes credit for linear functions answers — sharpened from recent examiner reports for the 2026 IB MYP Mathematics (Extended) sitting.

Gradient (slope)
Examiner keyword
The steepness of a line, $m = \dfrac{\Delta y}{\Delta x}$ .
y-intercept
Examiner keyword
The $y$ -coordinate of the point where a line crosses the y-axis (where $x = 0$ ).
Point-gradient form
Examiner keyword
$y - y_1 = m(x - x_1)$ — the equation of a line given its gradient and a point.

Common Mistakes and Misconceptions — Linear Functions

The traps other students keep falling into on linear functions questions — taken from recent IB MYP Mathematics (Extended) examiner reports and mark schemes — and how to avoid them.

✕Computing gradient as $\dfrac{\Delta x}{\Delta y}$ .
Why it happens
Mixing up rise and run.
How to avoid it
Gradient = RISE/RUN = $\dfrac{y_2 - y_1}{x_2 - x_1}$ . Y on TOP.
✕Writing perpendicular gradient as $\dfrac{1}{m_1}$ (no negative).
Why it happens
Missing the sign flip.
How to avoid it
Perpendicular = NEGATIVE RECIPROCAL. $m_1 \cdot m_2 = -1$ requires both: flip AND negate.
✕Calling two lines with same $c$ but different $m$ parallel.
Why it happens
Confusing 'crossing axis at the same place' with 'parallel'.
How to avoid it
Parallel requires SAME GRADIENT. y-intercept can be anything.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

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Linear Functions — frequently asked questions

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Linear Functions

Short Study Notes in the form of Slides

Short Study Notes — Linear Functions

Detailed Notes

What you’ll learn

How it’s examined

1Find gradient from two points

2Equation from gradient + point

3Equation from two points

4Perpendicular line

Gradient (slope)

y-intercept

Point-gradient form

✕Computing gradient as ΔxΔy\dfrac{\Delta x}{\Delta y}ΔyΔx​.

✕Writing perpendicular gradient as 1m1\dfrac{1}{m_1}m1​1​ (no negative).

✕Calling two lines with same ccc but different mmm parallel.

Practice questions

Past paper style quiz

Assess your understanding

Linear Functions — frequently asked questions

✕Computing gradient as $\dfrac{\Delta x}{\Delta y}$ .

✕Writing perpendicular gradient as $\dfrac{1}{m_1}$ (no negative).

✕Calling two lines with same $c$ but different $m$ parallel.