Edexcel · GCSE Maths · 1MA1 · Foundation and Higher

M12 · Coordinates and straight-line graphs

Revision notes, worked examples and methods for coordinates and straight-line graphs.

Revision notes ready · Quizzes and videos coming soon.

Revise the key ideas

Coordinates and gradients

  • Coordinates (x, y) give horizontal position first, then vertical position. In quadrant II, x is negative and y is positive; the origin is (0, 0).
  • The midpoint of (x₁, y₁) and (x₂, y₂) is (, ). The midpoint of (−2, 3) and (6, 7) is (2, 5).
  • A straight line has constant gradient m = . Use two well-separated points on the line; a positive gradient rises to the right.
    Graph of y equals two x plus oneThe line passes through zero one and two five. A gradient triangle rises four for a run of two.-101234-113579xyrun 2rise 4
    Graph of y equals two x plus one
  • Worked example: Through (1, 3) and (4, 9), m = = 2. Subtract the coordinates in the same order in numerator and denominator.
  • In y = mx + c, m is the gradient and c is the y-intercept. y = 2x + 1 passes through (0, 1), and goes up 2 for every 1 across.
  • To plot a line, calculate a table of values and draw a straight line through the points. Check the intercept and a third point to catch errors.
  • A horizontal line is y = k and has gradient 0. A vertical line is x = k; its gradient is undefined because the horizontal change is zero.

Finding line equations

  • Worked example: A line with gradient 3 through (2, 5) has 5 = 3 × 2 + c, so c = −1 and its equation is y = 3x − 1.
  • For a line through two points, find the gradient first, then substitute either point to find c. Through (1, 3) and (4, 9), y = 2x + 1.
  • Parallel non-vertical lines have equal gradients. y = 2x + 1 and y = 2x − 5 are parallel; different intercepts make them distinct lines.
  • Rearrange another line form to identify gradient: 2y = 6x − 4 becomes y = 3x − 2, so the gradient is 3, not 6.

Higher — perpendicular lines

  • For perpendicular non-horizontal, non-vertical lines, gradients satisfy m₁m₂ = −1. Take the negative reciprocal: the line perpendicular to gradient 2 has gradient −.
  • Worked example: Perpendicular to y = 2x + 1 through (4, 3): y = −x + c. Substitution gives 3 = −2 + c, so y = −x + 5.
  • A horizontal line is perpendicular to a vertical line. The negative-reciprocal rule should not be used by trying to divide by a zero gradient.

Test yourself

Quiz coming soon

Practice questions with explained answers will be added here.

For now, cover the worked answers, try the calculations yourself, then compare each step. Include units and reasons where needed.

Revision video

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