Linear GraphsCambridge IGCSE Maths: Revision notes

A straight-line graph has an equation in the form: $y = mx + c$ where:

• $m$ is the gradient (steepness) of the line
• $c$ is the y-intercept (where the line crosses the $y$-axis, i.e. the value of $y$ when $x=0$)

Extended candidates should also be able to work with the form $ax + by = c$, rearranging it into $y=mx+c$ form to identify $m$ and $c$.

Example: Rearrange $2x + 3y = 12$ into $y=mx+c$ form: $3y = -2x+12 \Rightarrow y = -\frac{2}{3}x + 4$, so $m=-\frac{2}{3}$, $c=4$.

Given two points $(x_1, y_1)$ and $(x_2, y_2)$ on a line, the gradient is: $m = \frac{y_2-y_1}{x_2-x_1}$

This is the change in $y$ divided by the change in $x$ between the two points.

Example: Find the gradient of the line through $(1, 3)$ and $(4, 9)$. $m = \frac{9-3}{4-1} = \frac{6}{3} = 2$

To find the equation of a line given two points, or a point and gradient:

1. Find the gradient $m$ (using the formula above, if not already given)
2. Substitute a known point $(x_1, y_1)$ and $m$ into $y - y_1 = m(x - x_1)$, or substitute into $y=mx+c$ and solve for $c$
3. Write the final equation in the form $y = mx + c$

Example: Find the equation of the line through $(2, 7)$ with gradient $3$. $y = 3x + c, \quad 7 = 3(2) + c \Rightarrow c = 1, \quad y = 3x + 1$

Parallel lines have exactly the same gradient.

Perpendicular lines meet at a right angle; their gradients multiply to give $-1$: $m_1 \times m_2 = -1 \quad \Rightarrow \quad m_2 = -\frac{1}{m_1}$

Example (parallel): Find the equation of the line parallel to $y=4x-1$ through $(1,-3)$. Same gradient $m=4$: $-3 = 4(1)+c \Rightarrow c=-7$, so $y=4x-7$.

Example (perpendicular): Find the gradient of a line perpendicular to a line with gradient $2$. $m_2 = -\frac{1}{2}$.

The perpendicular bisector of a line segment is the line that is perpendicular to it and passes through its midpoint — found by combining the midpoint formula with the perpendicular gradient rule.