Key Idea: Straight lines turn a gradient and a point into an equation you can graph, intersect or compare. They run through coordinate geometry on Paper 1 (by hand) — and feed straight into tangents and normals later in calculus.
📈 Gradient & the three forms
- first point on the line
- second point — subtract in the same order top and bottom
| Form | Looks like | Use it when… |
|---|---|---|
| Gradient–intercept | y = mx + c | you want gradient m and y-intercept c at a glance |
| Point–gradient | y − y₁ = m(x − x₁) | you're building a line from a point and a gradient |
| General | ax + by + d = 0 | the answer must be tidy with integer coefficients |
m > 0 uphill, m < 0 downhill, m = 0 horizontal (y = c); a vertical line x = a has no gradient (run = 0). From ax + by + d = 0 the gradient is m = −a⁄b — rearrange to y = mx + c. e.g. 3x + 2y − 6 = 0 → m = −1.5.
⟂ Parallel & perpendicular
| Relationship | Gradient rule | Example (from m = ⅔) |
|---|---|---|
| Parallel (never meet) | equal gradients, m₁ = m₂ | parallel gradient = ⅔ |
| Perpendicular (meet at 90°) | m₁ × m₂ = −1, so m₂ = −1⁄m₁ | flip & change sign → −3⁄2 |
For perpendicular, flip the fraction and swap the sign: ⅔ → −3⁄2. Treat a whole number as over 1: 5 = 5⁄1, so its perpendicular gradient is −1⁄5. A perpendicular bisector uses this gradient through the midpoint, $\left(\tfrac{x₁+ₓ₂}{2}, \tfrac{y₁+y₂}{2}\right)$.
✏️ IB-style worked examples
IB-style question — line through two points
Find the equation of the line through P(1, 2) and Q(3, 8). Give your answer as y = mx + c.
Step by step:
Gradient first — subtract in the same order top and bottom.
Use point–gradient with either point, say (1, 2).
Tidy.
y = 3x − 1.
IB-style question — where a line crosses the axes
Find where the line y = 2x − 6 crosses each axis.
Step by step:
y-intercept: put x = 0.
x-intercept: put y = 0 and solve.
Crosses the y-axis at (0, −6) and the x-axis at (3, 0).
IB-style question — line perpendicular through a point
Find the equation of the line through (4, 1) perpendicular to y = 2x − 3.
Step by step:
Perpendicular ⇒ negative reciprocal of 2.
Point–gradient form with (4, 1).
Tidy.
y = −½x + 3.
IB-style question — perpendicular bisector
Find the perpendicular bisector of A(1, 2) and B(5, 8).
Step by step:
Midpoint of AB — average the coordinates.
Gradient of AB, then take its negative reciprocal.
Point–gradient through the midpoint (3, 5).
y = −⅔x + 7.
Important: For perpendicular you need the negative reciprocal: flip the fraction and change the sign. From m = 2 the perpendicular gradient is −½, not −2 and not ½. And don't mix the rules — parallel keeps the gradient, perpendicular flips it.
Tap each card to reveal the answer.
Gradient of the line through (1, 2) and (4, 11) m = 3 — (11 − 2) ÷ (4 − 1) = 9⁄3.
Gradient of 4x + 2y − 8 = 0 m = −2 — rearrange to y = −2x + 4 (or m = −a⁄b = −4⁄2).
Line through (2, 5) with gradient 3 y = 3x − 1 — y − 5 = 3(x − 2), then tidy.
Gradient perpendicular to a line with gradient ⅔ −3⁄2 — flip the fraction and change the sign.
Midpoint of (1, 2) and (5, 8) (3, 5) — average the x's and the y's.
Exam Tips
- Gradient = rise ÷ run; subtract the points in the same order top and bottom.
- Building a line? Find the gradient, then drop it and a point into y − y₁ = m(x − x₁).
- From ax + by + d = 0 the gradient is m = −a⁄b — rearrange to y = mx + c.
- Parallel → equal gradients; perpendicular → negative reciprocal (m₁m₂ = −1).
- Perpendicular bisector = through the midpoint, with the negative-reciprocal gradient.