Linear Graphs (y = mx + c) (Edexcel IGCSE Maths): Revision notes

Section 1

What does y = mx + c actually mean?

Every straight line can be written as $y = mx + c$ where:

$m$ is the gradient (how steep the line is, and whether it slopes up or down)
$c$ is the y-intercept (where the line crosses the y-axis, at the point $(0, c)$ )

If $x$ increases by 1, $y$ changes by $m$ . A positive $m$ means the line slopes upward left to right; a negative $m$ means it slopes downward.

Always rearrange an equation into this form first before reading off $m$ and $c$ — do not assume the coefficient of $x$ is the gradient until the equation is arranged with $y$ on its own.

Key termsgradienty-intercept

Common mistake

Given $2y = 6x + 4$ , students often say $m = 6$ . Divide by 2 first: $y = 3x + 2$ , so $m = 3$ .

Section 2

How do I find the gradient between two points?

Given two points $(x_1, y_1)$ and $(x_2, y_2)$ , the gradient is:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

This is often remembered as 'change in y over change in x', or 'rise over run'. It does not matter which point you call point 1 and which you call point 2, as long as you are consistent — subtract in the same order on both the top and bottom.

Key termsrise over run

Example

Find the gradient through $(1, 4)$ and $(3, 10)$ : $m = \frac{10 - 4}{3 - 1} = \frac{6}{2} = 3$ .

Common mistake

Mixing the order, e.g. $\frac{y_1 - y_2}{x_2 - x_1}$ , flips the sign of the gradient. Keep the subtraction order matched on top and bottom.

Section 3

How do I find the full equation of a line from two points?

Steps:

Find the gradient $m$ using $\frac{y_2-y_1}{x_2-x_1}$ .
Substitute $m$ and one point $(x_1, y_1)$ into $y = mx + c$ .
Solve for $c$ .
Write the final equation in the form $y = mx + c$ .

Alternatively, use the point-gradient form directly:

$y - y_1 = m(x - x_1)$

then expand and rearrange into $y = mx + c$ form.

Key termspoint-gradient form

Example

Line through $(2, 5)$ and $(4, 11)$ : $m = \frac{11-5}{4-2} = 3$ . Using $(2,5)$ : $5 = 3(2) + c \Rightarrow c = -1$ . Equation: $y = 3x - 1$ .

Exam tip

Always check your answer by substituting the SECOND point into your final equation — it must also satisfy it.

Section 4

How do I tell if two lines are parallel or perpendicular?

Parallel lines have the exact same gradient: if $y = 3x + 2$ and $y = 3x - 5$ , both have $m = 3$ , so they are parallel and never meet.

Perpendicular lines meet at a right angle. Their gradients are negative reciprocals of each other:

$m_1 \times m_2 = -1 \quad \text{so} \quad m_2 = -\frac{1}{m_1}$

To find a perpendicular gradient: flip the fraction and change the sign. If $m_1 = 2$ (i.e. $\frac{2}{1}$ ), the perpendicular gradient is $-\frac{1}{2}$ . If $m_1 = -\frac{3}{4}$ , the perpendicular gradient is $\frac{4}{3}$ .

Key termsparallelperpendicularnegative reciprocal

Think of it like this

Think of gradients as directions on a slope: parallel lines climb at the identical angle side by side; perpendicular lines cut directly across each other, like a ladder resting against a wall meeting the flat ground.

Common mistake

Forgetting to flip AND change sign — students often do only one of the two steps when finding a perpendicular gradient.

Section 5

How do vertical and horizontal lines fit in?

Two special cases don't fit neatly into $y = mx + c$ :

Horizontal lines: $y = c$ (a constant). Gradient is 0.
Vertical lines: $x = c$ (a constant). Gradient is undefined (infinite steepness).

A vertical line is perpendicular to any horizontal line, and vice versa, even though the negative reciprocal rule cannot be applied numerically here (since you cannot divide by 0).

Key termshorizontal linevertical line

Exam tip

If a question gives you $x = 4$ , do not try to write it as $y = mx + c$ — it simply has no y-intercept form.

Must Know

$y = mx + c$ : $m$ is gradient, $c$ is y-intercept (the point $(0,c)$ )
Gradient formula: $m = \dfrac{y_2-y_1}{x_2-x_1}$
To find an equation from two points: find $m$ , then substitute one point to find $c$
Parallel lines: same gradient ( $m_1 = m_2$ )
Perpendicular lines: gradients multiply to $-1$ ( $m_2 = -\dfrac{1}{m_1}$ )
$y = c$ is horizontal (gradient 0); $x = c$ is vertical (gradient undefined)

That's the notes covered.

Carry on to the next subtopic.