Learn4MA1Papers 1H, 2H · Higher tierSequences, functions and graphsParallel and perpendicular lines

Sequences, functions and graphs · Subtopic 4 of 6

Parallel and Perpendicular Lines

Parallel lines have equal gradients. Perpendicular lines have gradients that multiply to $-1$ : flip the fraction and change the sign. With these two facts, plus one point, you can write down any related line.

Cambridge 4024 Paper 1Cambridge 4024 Paper 2Edexcel 4MA1 Papers 1H, 2HEdexcel 4MA1 Papers 1F, 2FAQA 8300 Papers 1–3≈ 30 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Know that parallel lines have equal gradients.
Find the gradient of a perpendicular line using $m_1m_2=-1$ .
Find the equation of a parallel or perpendicular line through a given point.
Find the perpendicular bisector of a line segment.

Pause and think

Rotate a line of gradient $2$ (up $2$ , across $1$ ) through $90^\circ$ . What happens to "up $2$ , across $1$ "?

Section 2 of 14

Core idea

Start hereRailway tracks and street corners

Railway tracks never meet: same direction, same gradient. Streets meeting at a corner turn through a right angle, and their gradients are linked in a surprising way: they multiply to $-1$ .

The big idea

Parallel: equal gradients. Perpendicular: $m_1m_2=-1$ , so $m_2=-\frac1{m_1}$ .

1. The rules

y = 2x + 1, a parallel line, and a perpendicular line

2\times\left(-\tfrac12\right)=-1

. The perpendicular line meets

y=2x+1

at a right angle.

Why perpendicular gradients multiply to −1

$\text{direction }(1,m)$
A line of gradient $m$ : right 1, up $m$ .
$\text{rotate }90^\circ:\ (-m,1)$
Rotating $(a,b)$ by $90^\circ$ gives $(-b,a)$ .
$\text{gradient}=\frac1{-m}=-\frac1m$
Rise over run.

Worked exampleParallel to $y=3x-1$ through $(2,8)$

Equation.

Goal: Line.
Strategy: Same gradient; substitute.

$m=3$
Parallel.
$8=6+c\Rightarrow c=2$
Point.

Answer

y=3x+2

✓ Check

3(2)+2=8

✓.

Worked examplePerpendicular to $y=2x+1$ through $(4,1)$

Equation.

Goal: Line.
Strategy: Negative reciprocal.

$m=-\tfrac12$
$-1\div2$ .
$1=-2+c\Rightarrow c=3$
Point.

Answer

y=-\frac12x+3

✓ Check

2\times(-\frac12)=-1

✓.

Common misconceptionOnly the sign Perpendicular to

m=2

-\frac12

, not

-2

Edge caseHorizontal and vertical

y=3

(gradient 0) and

x=2

(undefined) are perpendicular; the product rule doesn't apply.

Pause and think

Are $y=4x$ and $y=-\frac14x+2$ perpendicular?

Summary

Parallel: same $m$ .
Perpendicular: $-1/m$ .

Perpendicular: At right angles.

HistoryEuclid's fifth postulate. Euclid's Elements (about

300

BC) assumed that, through a point not on a line, exactly one parallel line can be drawn. For

2000

years mathematicians tried to prove it from his other rules. In the 1820s Lobachevsky and Bolyai showed it cannot be: there are consistent "non-Euclidean" geometries where it fails, and one of them describes the curved space of Einstein's relativity.

Fun factRailway tracks meet at the horizon. Parallel railway lines appear to meet in the distance. Renaissance painters used this "vanishing point" deliberately to make flat paintings look three-dimensional; the mathematics of perspective was worked out by artists like Brunelleschi in the 1400s.

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