Pearson Edexcel · GCSE · Mathematics
Course version: 1MA1 · Version from Mar 2026
Algebra - Graphs: plot graphs of equations that correspond to straight-line graphs in the coordinate plane; use the form y = mx + c to identify parallel and perpendicular lines; find the equation of the line through two given points or through one point with a given gradient (Higher tier)
Higher tier only; the addition to shared content is perpendicular lines.
Study notes
From the checked teaching notes for this specification point.
The idea
Higher tier only; the addition to shared content is perpendicular lines. In y = mx + c, m is the gradient and c the y-intercept. Parallel lines share the same gradient; perpendicular lines have gradients multiplying to −1, so each is the negative reciprocal of the other. A vertical line has no gradient and is perpendicular to a horizontal one. Rearrange so y stands alone before comparing: 3y = 6x + 5 has gradient 2, not 6. To plot a straight-line graph, substitute two or three x-values, plot the points and join them with a ruler. Through two points, find the gradient as the change in y over the change in x, then substitute one point to get c. For y = 4x − 1 through (8, 5), the perpendicular gradient is −1/4, and 5 = −2 + c gives c = 7, so y = −x/4 + 7.
What earns the marks
- Plotting a straight-line graph by substituting x-values, plotting the points and joining them with a ruled line.
- Reading the gradient from y = mx + c only after rearranging so that y is on its own.
- Identifying parallel lines by equal gradients, and perpendicular lines by gradients multiplying to −1.
- Using the negative reciprocal as the gradient of a perpendicular line, even if later work is wrong.
- Finding the gradient through two points as the change in y divided by the change in x, then substituting one point to find c.
- The finished equation, written in whichever form the question asks for.
Common mistakes and how to put them right
Turning the gradient upside down but leaving its sign alone, so pairing gradient 5 with 1/5 instead of −1/5.
Changing the sign but not turning the gradient upside down, so pairing gradient 5 with −5.
Reading the gradient straight off an equation such as 2y = 8x + 3 without dividing through first.
Keeping the intercept of the original line instead of substituting the point the question gives.
Joining plotted points freehand or with too few points, so the line is not straight.
Exam tips
- Write the perpendicular gradient as a fraction and cancel it, so −4/8 becomes −1/2 before you substitute.
- Check the pair by multiplying the two gradients together, because anything other than −1 means one of them is wrong.
- If a particular form of the equation is asked for, rearrange as the very last step so the method marks are already safe.
Try the lesson
The start of the Algebra - Graphs: plot graphs of equations that correspond to straight-line graphs in the coordinate plane; use the form y = mx + c to identify parallel and perpendicular lines; find the equation of the line through two given points or through one point with a given gradient (Higher tier) lesson, made from checked teaching notes for this specification point.
Have a go first
A straight line passes through the points (2, 7) and (6, 19). Find the equation of the line in the form y = mx + c.
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Test your understanding of Algebra - Graphs: plot graphs of equations that correspond to straight-line graphs in the coordinate plane; use the form y = mx + c to identify parallel and perpendicular lines; find the equation of the line through two given points or through one point with a given gradient (Higher tier)
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