Library Straight-Line Graphs
Additional Mathematics

Straight-Line Graphs

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Cambridge (CIE) IGCSE · Additional Maths

Straight-Line Graphs

Big idea: Every straight line has exactly one "steepness" (gradient) and one starting height (y-intercept) — once you know two points, or one point and the steepness, you can write down the line's entire equation and answer almost any question about it.

OverviewEverything in this chapter, at a glance
  • The gradient m of a line measures its slope: m = (y₂−y₁)/(x₂−x₁)
  • Three ways to write a line's equation: y = mx + c (gradient-intercept), y − y₁ = m(x − x₁) (point-gradient), and ax + by + d = 0 (general form)
  • The midpoint of two points is just the average of their x's and the average of their y's
  • The length of a line segment comes from Pythagoras: d = √[(x₁−x₂)² + (y₁−y₂)²]
  • Parallel lines have the exact same gradient (m₁ = m₂)
  • Perpendicular lines have gradients that multiply to −1 (m₁ × m₂ = −1) — i.e. they're negative reciprocals of each other
Topic 1Finding the Gradient of a Straight Line

What "gradient" actually means

Think of the gradient as the answer to the question: "For every step I take to the right, how much does the line go up (or down)?" It's a single number that completely describes how steep and in which direction a line tilts.

A gradient of 1 means: move 1 unit right → the line rises 1 unit. A gradient of −2 means: move 1 unit right → the line falls 2 units (the negative sign tells you it's going downhill as you move right). The bigger the number (ignoring the sign), the steeper the line. A gradient of 0 means the line is perfectly flat (horizontal).

y | * (x2, y2) | .·" | .·" | | .·" | dy (change in y) | .·"________| | .·" dx | .·" (change in x) |·* (x1, y1) +------------------------------ x
m = (y₂ − y₁) / (x₂ − x₁)
gradient = (difference in y) ÷ (difference in x) — often remembered as "rise over run"
⚡ Why this formula works It's literally just the definition of slope from real life — how much height you gain (dy) for how much distance you travel horizontally (dx). It doesn't matter which point you call (x₁,y₁) and which you call (x₂,y₂), as long as you're consistent — subtract in the same order top and bottom.
Worked Example
Find the gradient of the line through (−2, 6) and (8, 1).
m = (6 − 1) / (−2 − 8)
m = 5 / (−10)
Practice Question 1
Find the gradient of the line passing through (3, −1) and (7, 9).
Practice Question 2
A line passes through (−4, 5) and (2, 5). What is its gradient, and what does this tell you about the line?
Topic 2The Three Equations of a Straight Line

  y = mx + c — Gradient-Intercept Form

This is the most useful form because you can read the gradient and y-intercept straight off it — no rearranging needed. Here, m is the gradient, and c is the y-intercept, meaning the line crosses the y-axis at the point (0, c).

y = mx + c
m = gradient · c = value of y when x = 0 (where the line crosses the y-axis)

  y − y₁ = m(x − x₁) — Point-Gradient Form

This form is your best friend when you're building an equation from scratch, because it only needs a gradient and a single point on the line — you plug them straight in. Once you've written it, you can rearrange into y = mx + c or ax + by + d = 0 if the question asks for a specific format.

y − y₁ = m(x − x₁)
m = gradient · (x₁, y₁) = any known point on the line
💡 The strategy examiners expect Whenever you're asked to "find the equation of the line...", start with point-gradient form. Step 1: get the gradient (either given, or calculated from two points). Step 2: substitute in one known point. Step 3: rearrange into whatever form the question wants.

  ax + by + d = 0 — General Form

This is the "tidy, no-fractions" form examiners often ask for explicitly, especially when a, b and d must be integers. Its main superpower: you can instantly read off both intercepts without rearranging into y = mx + c first.

ax + by + d = 0
x-intercept: (−d/a, 0)   |   y-intercept: (0, −d/b)

Finding an equation from scratch — the full method

Worked Example (a) — Gradient + one point given
Find the equation of the line with gradient 3 that passes through (5, 4).
Since the gradient is 3, the line has the form:
y = 3x + c
Substitute (5, 4) to find c:
4 = 3(5) + c
4 = 15 + c  →  c = −11
Worked Example (b) — Two points given
Find the equation of the line passing through (−2, 6) and (8, 1).
Step 1 — find the gradient:
m = (6 − 1)/(−2 − 8) = 5/(−10) = −1/2
Step 2 — write y = −½x + c, then substitute the "nicer" point (8, 1) to avoid extra negative signs:
1 = −½(8) + c  →  1 = −4 + c  →  c = 5
✓ Sanity check: negative gradient (line goes downhill) and crosses the y-axis between 1 and 6 — matches a quick sketch.
🎯 Examiner tip Quickly sketching the line can save you from silly errors — if your final equation doesn't match the shape you'd expect from the sketch (uphill vs downhill, where it crosses the axes), go back and check your algebra. Also always check the question for the required output format: "usually y = mx + c or ax + by + d = 0", and check whether coefficients must be integers (they usually must for ax + by + d = 0).
Practice Question 3
Find the equation of the line with gradient −4 that passes through (1, 2). Give your answer in the form y = mx + c.
Practice Question 4
Find the equation of the line passing through (0, −3) and (4, 5). Give your answer in the form ax + by + d = 0 with integer coefficients.
Topic 3Midpoint of a Line Segment

Just an average of the coordinates

The midpoint is exactly what it sounds like: the point sitting exactly halfway between two endpoints, the same distance from each. There's no clever trick here beyond averaging — average the x-coordinates to get the midpoint's x, and average the y-coordinates to get the midpoint's y.

M = ( (x₁+x₂)/2 , (y₁+y₂)/2 )
Average the x's, average the y's — that's your midpoint
Worked Example
A is (−4, 3) and B is (8, −12). Find M, the midpoint of AB.
M = ( (−4+8)/2 , (3+(−12))/2 )
M = ( 4/2 , −9/2 )
Practice Question 5
Find the midpoint of the line segment joining (−6, 2) and (10, −8).
Topic 4Length of a Line Segment

It's really just Pythagoras in disguise

Picture the line segment as the hypotenuse of a right-angled triangle, where one leg is the horizontal distance between the points (the difference in x) and the other leg is the vertical distance (the difference in y). Pythagoras' theorem (a² = b² + c²) then gives you the length directly.

(x1,y1) * | \ | \ diff in| \ d (the length we want) y-coord| \ |_________\ * (x2,y2) diff in x-coord
d = √[(x₁ − x₂)² + (y₁ − y₂)²]
This is Pythagoras' theorem applied to the horizontal and vertical gaps between the two points
Worked Example
A is (3, −4) and B is (−5, 2). Calculate the distance AB.
d = √[(3 − (−5))² + (−4 − 2)²]
d = √[(8)² + (−6)²] = √[64 + 36] = √100
🧮 Calculator strategy Work with the square of the distance for as long as possible — this avoids early rounding errors. On the non-calculator paper, you may need to leave your answer as a surd (e.g. √52 rather than 7.21). Only take the square root right at the end, or when the question forces you to.
Practice Question 6
Point P has coordinates (1, 1) and point Q has coordinates (4, 5). Calculate the distance PQ.
Topic 5Parallel Lines

Same steepness, never meet

Parallel lines are always the same distance apart, no matter how far you follow them — which means they never intersect. The reason this works geometrically is simple: if two lines tilt at exactly the same angle, they can never converge or diverge. That "same angle" condition translates mathematically into "same gradient."

l₁ ∥ l₂  ⇔  m₁ = m₂
Two lines are parallel if and only if their gradients are equal

How to check: rearrange each line into y = mx + c form, then compare the coefficients of x. If they match, the lines are parallel.

Worked Example
Find the equation of the line parallel to y = 3x + 7 that passes through (2, 1).
Same gradient (3), so the line has the form:
y = 3x + d
Substitute (2, 1):
1 = 3(2) + d → 1 = 6 + d → d = −5
👀 Watch for hidden parallel lines Exam questions love to disguise parallel lines with phrases like "...at the same rate as..." or by describing the sides of a parallelogram/rectangle. Always check the properties of any geometric shape mentioned in the question — opposite sides of a parallelogram are automatically parallel.
Practice Question 7
Find the equation of the line parallel to 2x + y − 5 = 0 that passes through (3, −1). Give your answer in the form y = mx + c.
Topic 6Perpendicular Lines

Meeting at a perfect right angle

Perpendicular lines cross each other at exactly 90°. The relationship between their gradients is less obvious than for parallel lines, but it's a rule worth just memorising: multiply the two gradients together and you always get −1. Equivalently, one gradient is the negative reciprocal of the other — flip the fraction upside down, and flip the sign.

l₁ ⊥ l₂  ⇔  m₁ × m₂ = −1
e.g. if one gradient is 2, the perpendicular gradient is −1/2. If one gradient is −3/4, the perpendicular gradient is 4/3.
⚠️ Special case — horizontal & vertical lines The formula m₁ × m₂ = −1 breaks down for vertical lines (their gradient is undefined). Instead, remember directly: x = p and y = q are always perpendicular to each other (for any constants p, q) — a vertical line is always perpendicular to a horizontal line, by definition, no formula needed.
Worked Example
Find the equation of the line perpendicular to y = 2x − 2 that passes through (2, −3). Leave your answer in the form ax + by + c = 0 with integer coefficients.
The given line has gradient m₁ = 2, so the perpendicular gradient is the negative reciprocal:
m₂ = −1/2
Substitute (2, −3) into y = −½x + c:
−3 = −½(2) + c → −3 = −1 + c → c = −2
So the line is y = −½x − 2. Now rearrange to remove the fraction and match the required form:
y + ½x + 2 = 0  →  2y + x + 4 = 0
🎯 Examiner tip Exam questions are excellent at "hiding" perpendicularity. Classic examples: a tangent and a radius of a circle are always perpendicular; the diagonals of certain shapes are perpendicular; adjacent sides of a rectangle/square are perpendicular. Train yourself to spot these geometric clues.
Practice Question 8
Find the equation of the line perpendicular to y = −4x + 1 that passes through (8, 3). Give your answer in the form y = mx + c.
Practice Question 9
A line l has equation 3x + y − 6 = 0. Is the line y = ⅓x + 4 parallel to l, perpendicular to l, or neither?
Quick ReferenceWhat to Memorise
Gradient formula
m = (y₂ − y₁)/(x₂ − x₁) — rise over run between any two points on the line
Gradient-intercept form
y = mx + c — read gradient and y-intercept directly
Point-gradient form
y − y₁ = m(x − x₁) — best starting point when building an equation
General form
ax + by + d = 0 — x-intercept (−d/a, 0), y-intercept (0, −d/b)
Midpoint formula
M = ((x₁+x₂)/2, (y₁+y₂)/2) — average the coordinates
Length (distance) formula
d = √[(x₁−x₂)² + (y₁−y₂)²] — Pythagoras applied to coordinate differences
Parallel lines condition
m₁ = m₂ — identical gradients, never intersect
Perpendicular lines condition
m₁ × m₂ = −1 — negative reciprocal gradients, cross at 90°
Vertical/horizontal special case
x = p and y = q are always perpendicular to each other, for any constants p, q
Self-checkConcepts Checklist
Final PrepExam Tips & Common Mistakes

Mistakes students actually make

  • Subtracting coordinates in the wrong order — if you do (y₂−y₁) on top, you MUST do (x₂−x₁) on the bottom, using the same point order both times, or your gradient's sign flips.
  • Forgetting to flip AND negate for perpendicular gradients — going from gradient 2 to gradient 2 (forgot to flip) or to gradient −2 (forgot to reciprocate) instead of the correct −1/2.
  • Leaving fractions in "integer coefficient" answers — if a question asks for ax + by + d = 0 with integers, always multiply through to clear any fractions before giving your final answer.
  • Rounding too early in distance calculations — always keep the exact square root (or square) until the very last step.
  • Assuming lines with similar-looking equations are parallel without actually rearranging both into y = mx + c form first to properly compare gradients.
  • Missing hidden geometric clues — not recognising that "the tangent to the circle" implies perpendicularity to the radius, or that a rectangle's sides imply both parallel and perpendicular relationships.

What examiners are looking for

  • Clear, methodical working — showing the gradient calculation as a separate step before substituting into point-gradient form.
  • Final answers given in the exact format requested in the question (check for "in the form y = mx + c" vs "in the form ax + by + c = 0" instructions).
  • Integer coefficients when the general form is requested — don't leave halves or thirds floating around.
  • A quick sketch as a sanity check on multi-step geometry questions — it costs no marks and catches sign errors instantly.
🚀 Last-minute strategy If you get stuck on any "find the equation of the line..." question, always default to this 3-step method: (1) Find the gradient. (2) Plug gradient + one point into y − y₁ = m(x − x₁). (3) Rearrange to the requested format. This single method solves nearly every question type in this chapter.
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