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.
Notes and quizzes ready · 50 questions · Video coming soon.
Foundation shows shared notes and questions. Higher includes the shared content and labelled Higher extensions. Changing tier starts the notes and a new quiz set.
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 (x₁ + x₂2, y₁ + y₂2). The midpoint of (−2, 3) and (6, 7) is (2, 5).
A straight line has constant gradient m = change in ychange in x. Use two well-separated points on the line; a positive gradient rises to the right. Graph of y equals two x plus one
Worked example: Through (1, 3) and (4, 9), m = 9 − 34 − 1 = 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 −12.
Worked example: Perpendicular to y = 2x + 1 through (4, 3): y = −12x + c. Substitution gives 3 = −2 + c, so y = −12x + 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
50 questions · Sets of 10 from the selected tier. For fractions, use / when typing; for powers, use superscripts or ^. Follow each question's answer format. These quick checks support revision; practise full written solutions and proofs too.
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