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06062025–2027 syllabus

ADDITIONAL MATHEMATICS · CHAPTER 7

Straight-line graphs

Use coordinate geometry directly, then linearise nonlinear data so gradient and intercept reveal hidden constants.

Additional Mathematics4 connected sectionsSyllabus-aligned guide

LEARNING OBJECTIVES

What you will be able to do

  • use equivalent equations of a straight line
  • apply parallel and perpendicular gradient conditions
  • find lengths, midpoints and perpendicular bisectors
  • transform relationships to and from straight-line form

AT A GLANCE

Syllabus0606Coverage2025–2027Sections4LevelAdditional Mathematics

INTRODUCTION · THE BIG IDEA

Use coordinate geometry directly, then linearise nonlinear data so gradient and intercept reveal hidden constants.

Straight-line methods support coordinate geometry, tangents, normals and data modelling. The safest equation form depends on what information is given.

A nonlinear relationship can become linear after a deliberate change of plotted variables. The new axes determine what the gradient and intercept mean.

01

SECTION 01

Line equations and gradients

Core concept

Use y = mx + c when the intercept matters and y − y₁ = m(x − x₁) when a point and gradient are known. Vertical lines have equations x = constant and undefined gradient.

RULE 1
m = (y₂ − y₁)/(x₂ − x₁)
RULE 2
y − y₁ = m(x − x₁)
RULE 3
y = mx + c
Gradient triangle between two coordinate pointsA(x₁, y₁)B(x₂, y₂)run = x₂ − x₁rise = y₂ − y₁gradient = rise ÷ run
Read both coordinate differences in the same direction; reversing both leaves the gradient unchanged.
02

SECTION 02

Parallel, perpendicular and bisector problems

Core concept

Parallel non-vertical lines have equal gradients. Perpendicular finite non-zero gradients multiply to −1. A perpendicular bisector passes through the midpoint and has perpendicular gradient.

RULE 1
m_parallel = m
RULE 2
m₁m₂ = −1 for perpendicular lines
RULE 3
midpoint = ((x₁+x₂)/2, (y₁+y₂)/2)
RULE 4
length = √[(x₂−x₁)² + (y₂−y₁)²]
Original worked example

Construct a perpendicular bisector

  1. Take A(1, 2) and B(5, 4).
  2. Midpoint M = (3, 3).
  3. Gradient AB = (4 − 2)/(5 − 1) = 1/2, so perpendicular gradient = −2.
  4. Use point-gradient form through M.

Answer: y − 3 = −2(x − 3), or y = −2x + 9.

03

SECTION 03

Power and exponential linearisation

Core concept

For y = Axⁿ, taking logs produces log y = n log x + log A. For y = Abˣ, log y = x log b + log A. Match these to Y = mX + c using the actual plotted axes.

RULE 1
log y = n log x + log A
RULE 2
log y = x log b + log A
Common straight-line transformations
Original relationPlotGradientIntercept
y = Axⁿlog y against log xnlog A
y = Abˣlog y against xlog blog A
y² = Ax³ + By² against x³AB
y³ = A ln x + By³ against ln xAB
04

SECTION 04

Reading and rebuilding a model

Core concept

Label transformed coordinates, calculate gradient from two well-separated points on the fitted line, then translate m and c back into the original constants. The same process works in reverse when a straight-line graph is described first.

ORIGINAL STUDY DIAGRAMDecode a transformed graph
1Name X and Y axes
2Write Y = mX + c
3Match terms to the model
4Find transformed constants
5Undo powers or logs

QUICK CHAPTER SUMMARY

The ideas to carry forward

  • Choose a line equation that matches the given information.
  • Parallel gradients match; perpendicular gradients are negative reciprocals.
  • A perpendicular bisector uses both midpoint and gradient.
  • Logarithms linearise power and exponential laws.
  • Transform the intercept back before stating an original constant.

QUICK REVISION CHECKLIST

Can you do each of these without your notes?

  • use equivalent equations of a straight line
  • apply parallel and perpendicular gradient conditions
  • find lengths, midpoints and perpendicular bisectors
  • transform relationships to and from straight-line form
← Logarithmic and exponential functionsCoordinate geometry of the circle →