Straight-line Graphs and y = mx + c — AQA GCSE Maths Revision Notes

What you'll learn

This topic covers the equation of a straight line. By the end of this guide you should be able to plot a line from its equation, read the gradient and intercept from y = mx + c, and calculate a gradient from two points.

You should also be able to find the equation of a line through given points, rearrange an equation into the standard form, recognise parallel and perpendicular lines from their gradients, and find where two lines cross.

The organising idea is that y = mx + c holds the whole line in two numbers. The gradient m says how steep the line is and which way it tilts; the intercept c says where it crosses the y-axis. Those two facts fix the line completely — no other information is needed to draw it, and every question in this topic is either extracting them from something, or using them to produce something else. Getting an equation into that form is therefore almost always the first move, because until you do, neither number is visible.

Key terms and definitions

Gradient (m) — the steepness of the line: how far it rises for each unit across.

y-intercept (c) — the value of y where the line crosses the y-axis.

x-intercept — where the line crosses the x-axis, found by setting y = 0.

Linear — producing a straight line, with x to the first power only.

Parallel — having the same gradient.

Perpendicular — meeting at a right angle, with gradients whose product is −1.

Midpoint — the point halfway between two others.

Core concepts

Reading y = mx + c

In the equation y = mx + c, the number multiplying x is the gradient and the constant is the y-intercept.

So y = 2x + 3 has gradient 2 and crosses the y-axis at 3. And y = −4x + 1 has gradient −4 and intercept 1.

Watch for missing pieces. In y = 3x there is no constant, so c = 0 and the line passes through the origin. In y = 5 there is no x term, so the gradient is 0 and the line is horizontal.

Watch also for the order. In y = 7 − 2x the gradient is −2, not 7 — the number attached to x is the gradient wherever it appears, so rewriting it as y = −2x + 7 makes that clear.

Rearranging into the standard form

Equations are often given in other forms, and the gradient cannot be read until y is the subject.

For 2y = 6x + 4, divide everything by 2 to get y = 3x + 2, so the gradient is 3.

For 3x + y = 7, subtract 3x to get y = −3x + 7, so the gradient is −3.

For 2x + 4y = 12, subtract 2x and divide by 4 to get y = −0.5x + 3.

Reading a gradient straight off an unrearranged equation is one of the commonest errors, and rearranging first removes it entirely.

What the gradient means

A positive gradient rises from left to right; a negative gradient falls.

The size gives the steepness, so y = 5x is steeper than y = 2x, and y = −5x is equally steep but falling.

A gradient of 3 means the line rises 3 units for every 1 unit across. A gradient of ½ means it rises 1 unit for every 2 across.

Calculating the gradient from two points

The gradient is the change in y divided by the change in x.

For the points (1, 4) and (5, 16): the change in y is 16 − 4 = 12, and the change in x is 5 − 1 = 4, so the gradient is 12 ÷ 4 = 3.

Take the differences in the same order for both. Subtracting the y values one way and the x values the other gives the wrong sign, which turns a rising line into a falling one.

A negative gradient arises naturally from this: for (2, 9) and (6, 1), the change in y is 1 − 9 = −8 and the change in x is 4, giving a gradient of −2.

Finding the equation from a point and a gradient

Substitute the known point into y = mx + c and solve for c.

A line with gradient 3 through the point (2, 11): substituting gives 11 = 3(2) + c, so 11 = 6 + c and c = 5.

The equation is y = 3x + 5.

Checking with the original point confirms it: 3(2) + 5 = 11. ✓

Finding the equation from two points

Find the gradient first, then use either point to find c.

Through (1, 4) and (5, 16): the gradient is 3, as calculated above. Substituting (1, 4) gives 4 = 3(1) + c, so c = 1.

The equation is y = 3x + 1.

Using the other point as a check is worth the extra line: 3(5) + 1 = 16. ✓ If the two points give different values of c, the gradient was wrong.

Plotting a line

Two methods, and the right choice depends on the form of the equation.

From y = mx + c, mark the intercept on the y-axis, then use the gradient to step to a second point — along 1 and up m — and draw the line through them. Plotting a third point as a check is sensible, since three points in a line confirm the first two.

From an equation in any form, make a small table of values, substituting two or three x-values and calculating y for each.

For lines such as x = 3 or y = −2, no calculation is needed: x = a number is a vertical line and y = a number is horizontal. Confusing these two is common, and the way to settle it is that x = 3 means every point on the line has x-coordinate 3, which describes a vertical line.

Parallel and perpendicular lines

Parallel lines have the same gradient. So y = 4x + 1 and y = 4x − 7 are parallel, differing only in where they cross the axis.

Perpendicular lines have gradients whose product is −1, which means each is the negative reciprocal of the other.

So a line of gradient 2 is perpendicular to one of gradient −½, and a line of gradient −¾ is perpendicular to one of gradient 4/3.

Both steps are required: invert the fraction and change the sign. Doing only one is the usual error.

To find a line perpendicular to y = 2x + 5 through the point (4, 3): the new gradient is −½, and substituting gives 3 = −½(4) + c, so 3 = −2 + c and c = 5. The equation is y = −½x + 5.

Where two lines cross

The crossing point satisfies both equations, so it is found by solving them simultaneously.

For y = 2x + 1 and y = 5 − x: setting them equal gives 2x + 1 = 5 − x, so 3x = 4 and x = 4/3. Substituting back into either equation gives y = 11/3.

Graphically, this is the point where the drawn lines intersect, and reading it off a graph is an acceptable method when the question provides one.

Parallel lines never cross, which is why solving their equations simultaneously produces a contradiction rather than a solution.

Worked examples

Example 1: Rearranging and reading

Find the gradient and y-intercept of the line 4x + 2y = 10.

Rearrange to make y the subject. Subtract 4x: 2y = −4x + 10.

Divide everything by 2: y = −2x + 5.

The gradient is −2 and the y-intercept is 5.

Reading the original equation would have suggested a gradient of 4, which is why the rearrangement must come first.

Example 2: The equation through two points

Find the equation of the line through (−2, 7) and (4, −5).

Gradient = change in y ÷ change in x = (−5 − 7) ÷ (4 − (−2)) = −12 ÷ 6 = −2.

Substitute (4, −5) into y = mx + c: −5 = −2(4) + c, so −5 = −8 + c and c = 3.

The equation is y = −2x + 3.

Check with the other point: −2(−2) + 3 = 4 + 3 = 7. ✓ Both points satisfy it, so the answer is confirmed.

Example 3: A perpendicular line

Find the equation of the line perpendicular to y = 3x − 4 passing through (6, 1).

The given gradient is 3, so the perpendicular gradient is the negative reciprocal: −1/3.

Substitute the point: 1 = −1/3 × 6 + c, so 1 = −2 + c and c = 3.

The equation is y = −1/3 x + 3.

Check the gradients multiply to −1: 3 × (−1/3) = −1. ✓ Taking only the reciprocal, without the sign change, would have given 1/3 and a line that is not perpendicular at all.

Common mistakes and how to avoid them

Reading the gradient from an unrearranged equation. Make y the subject first.

Taking the gradient as the first number. In y = 7 − 2x the gradient is −2.

Subtracting the coordinates in inconsistent orders. Keep the same order for both differences.

Forgetting the sign change for a perpendicular gradient. Invert and negate.

Confusing x = 3 with y = 3. x = a number is vertical.

Plotting only two points. A third confirms them and catches an arithmetic slip.

Assuming parallel lines have the same equation. They share a gradient but differ in c.

Exam technique for "Straight-line Graphs"

Rearrange into y = mx + c as your first written step whenever the equation is given in another form. The gradient and intercept are then simply read off.

Write the gradient calculation as a fraction, with the y differences on top, so the method mark is visible even if the arithmetic slips.

Check an equation you have found by substituting both given points, not just the one you used.

For perpendicular gradients, confirm the two multiply to −1 before moving on.

Plot three points rather than two, and use a ruler.

State gradients and intercepts explicitly when asked to describe a line, rather than only giving the equation.

Quick revision summary

y = mx + c holds the whole line in two numbers: m is the gradient, c is the y-intercept. Rearrange into this form before reading either.

Gradient = change in y ÷ change in x, with both differences taken in the same order. Positive rises, negative falls, and the size is the steepness.

To find an equation, get the gradient first, then substitute a point to find c — and check with the second point.

x = a number is a vertical line; y = a number is horizontal.

Parallel lines have the same gradient. Perpendicular gradients multiply to −1, so each is the negative reciprocal of the other: 2 and −½. Invert and change the sign.

Two lines cross where their equations are satisfied simultaneously; parallel lines never do.

Plot from the intercept using the gradient as a step, and check with a third point.