Edexcel GCSE Maths Foundation (1MA1), Foundation tier · Algebra
Practise Straight-line graphs. unlimited generated questions on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Plotting and interpreting straight-line graphs: gradient, y-intercept, \(y = mx + c\), finding the equation of a line from two points or a point and a gradient, and parallel lines (both tiers). Higher also covers perpendicular lines.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
3
Plot a straight line from its equationMake a table of values, plot the points and join them with a ruler.
4
Find the gradient between two pointsGradient = change in y ÷ change in x, e.g. from (1, 2) to (4, 11): \(9 \div 3 = 3\).
4
Identify gradient and intercept from an equationIn \(y = 5x - 2\), the gradient is 5 and the line crosses the y-axis at (0, −2).
5
Find the equation of a lineWork out m, then substitute a point to find c, e.g. through (2, 7) and (4, 11) gives \(y = 2x + 3\).
5
Identify and find parallel linesParallel lines have equal gradients: \(y = 3x + 1\) is parallel to \(y = 3x - 4\).
5
Rearrange an equation to find the gradientMake y the subject, e.g. \(2y + 6x = 5\) gives \(y = -3x + 2.5\), so the gradient is −3.
5
Interpret gradient and intercept in contexte.g. in \(C = 45d + 30\), 45 is the cost per day and 30 is a fixed charge.
Notes
Gradient and intercept
The gradient m measures steepness: how far up the line goes for every 1 unit across.
Gradient \(= \frac{\text{change in } y}{\text{change in } x} = \frac{y_2 - y_1}{x_2 - x_1}\).
Lines going uphill from left to right have positive gradients; downhill lines have negative gradients; horizontal lines have gradient 0.
The y-intercept c is where the line crosses the y-axis, at \((0, c)\).
\(x = a\) is a vertical line and \(y = b\) is a horizontal line.
Using \(y = mx + c\)
Make y the subject before reading off m and c: \(3x + 2y = 8\) gives \(y = -\frac{3}{2}x + 4\).
From a point and a gradient: gradient 4 through (2, 5) gives \(5 = 4 \times 2 + c\), so \(c = -3\) and \(y = 4x - 3\).
From two points: through (1, 6) and (3, 2), \(m = \frac{2 - 6}{3 - 1} = -2\). Then \(6 = -2 + c\), so \(c = 8\) and \(y = -2x + 8\).
Parallel lines have the same gradient.
Straight lines in context
The gradient is a rate of change, e.g. pounds per hour or litres per minute.
The y-intercept is the starting value, e.g. a fixed charge before any hours are worked.
Cheatsheet
\(y = mx + c\): m is the gradient, c is the y-intercept
Gradient \(= \frac{y_2 - y_1}{x_2 - x_1}\)
Parallel lines: same gradient
\(x = a\) is vertical; \(y = b\) is horizontal
Midpoint of \((x_1, y_1)\) and \((x_2, y_2)\): \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\)
How to answer each type of question
Find the equation of a line through two points
3 marks5
Work out the gradient from the two points.
Substitute one point into \(y = mx + c\) to find c.
Write the equation and check it with the other point.
Example. Find the equation of the line that passes through (−1, 7) and (3, −5).