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.
Foundation shows shared notes. Higher includes shared notes and the labelled Higher extensions. Changing tier starts the notes again.
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
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.