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Mathematics 05803.1

Straight line graphs

Gradient, intercept, y = mx + c, and parallel and perpendicular lines.

Learning objectives

What you need to be able to do

Teacher-mapped phrasing — check against the official Cambridge syllabus for exact wording.

  • 3.1.1Find the gradient and equation of a straight line and interpret y = mx + c.
  • 3.1.2Determine the conditions for lines to be parallel or perpendicular.Supplement

6 minute read

Straight line graphs

The equation of a straight line

y = mx + c where m is the gradient and c is the y-intercept — where the line crosses the y-axis.

Gradient

m = (y₂ − y₁) / (x₂ − x₁) — the change in y divided by the change in x, often remembered as "rise over run".

  • A positive gradient slopes upward from left to right.
  • A negative gradient slopes downward.
  • A steeper line has a larger magnitude of gradient.

Take the coordinates in the same order on top and bottom. Swapping one but not the other flips the sign.

Finding the equation from two points

  1. Calculate the gradient m.
  2. Substitute m and one point into y = mx + c and solve for c.
  3. Write the full equation.

Parallel and perpendicular

  • Parallel lines have equal gradients: m₁ = m₂.
  • Perpendicular lines have gradients whose product is −1: m₁ × m₂ = −1, so m₂ is the negative reciprocal of m₁.

If a line has gradient 2, a perpendicular line has gradient −1/2. If it has gradient −3/4, a perpendicular line has gradient 4/3.

Midpoint and length

  • Midpoint = the mean of the coordinates: ((x₁+x₂)/2, (y₁+y₂)/2).
  • Length by Pythagoras: √((x₂−x₁)² + (y₂−y₁)²).

Think of it like this

The negative reciprocal makes sense physically: turning a slope through 90° swaps how far you go across with how far you go up, and reverses the direction — which is exactly "flip the fraction and change the sign".

Worked examples

Method, step by step

Find the equation of the line passing through (1, 5) and (3, 11).

  1. 1Gradient m = (11 − 5) / (3 − 1) = 6/2 = 3.
  2. 2Substitute m = 3 and the point (1, 5) into y = mx + c: 5 = 3(1) + c.
  3. 3So c = 2.
  4. 4The equation is y = 3x + 2. Check with (3, 11): 3(3) + 2 = 11 ✓

y = 3x + 2

A line has equation y = 4x − 7. Find the gradient of a line perpendicular to it.

  1. 1The gradient of the given line is m₁ = 4.
  2. 2Perpendicular gradients satisfy m₁ × m₂ = −1.
  3. 3So m₂ = −1/4.

−1/4

Common misconceptions

  • Calculating gradient as change in x over change in y. It is always **y over x**.
  • Saying perpendicular gradients are just negatives of each other. The gradient perpendicular to 2 is −1/2, not −2.
  • Reading the y-intercept off a graph without checking the scale on the axes.

In the exam

  • When finding a line through two points, always verify by substituting the *other* point at the end — it should satisfy the equation.
  • Rearrange any equation into y = mx + c form before comparing gradients; `2y = 6x + 4` has gradient 3, not 6.