Straight-Line Graphs

Straight-line graphs represent linear relationships between two variables. Understanding the equation of a straight line, its gradient, and its intercept is fundamental to AQA GCSE Mathematics. This topic connects algebra with geometry and appears on both Foundation and Higher tier papers.

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The Equation y = mx + c

Every straight line can be written in the form:

y = mx + c

where:

• m is the gradient (steepness) of the line.
• c is the y-intercept (where the line crosses the y-axis).

Feature	Meaning	Example in y = 3x - 2
m (gradient)	How steep the line is; rise over run	m = 3 (goes up 3 for every 1 across)
c (y-intercept)	Where the line crosses the y-axis	c = -2 (crosses at (0, -2))

Exam Tip: If a line equation is written as y = c + mx (the terms swapped), it is still in the same form. The coefficient of x is always the gradient and the constant term is always the y-intercept.

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Gradient

The gradient measures how steep a line is. It is calculated as:

Gradient = change in y / change in x = (y2 - y1) / (x2 - x1)

Gradient Type	Description	Visual
Positive gradient	Line slopes upward from left to right	/
Negative gradient	Line slopes downward from left to right	\
Zero gradient	Horizontal line	—
Undefined	Vertical line

Worked Example 1

Find the gradient of the line passing through (2, 3) and (6, 11).

Gradient = (11 - 3) / (6 - 2) = 8 / 4 = 2

Worked Example 2

Find the gradient of the line passing through (1, 7) and (4, 1).

Gradient = (1 - 7) / (4 - 1) = -6 / 3 = -2

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Plotting a Straight-Line Graph

To plot a straight line given an equation:

Method 1: Table of Values

1. Choose three x-values (three points are enough; the third is a check).
2. Substitute each into the equation to find y.
3. Plot the points and draw a straight line through them.

Worked Example 3

Plot: y = 2x + 1

x	y = 2x + 1	Point
0	1	(0, 1)
1	3	(1, 3)
3	7	(3, 7)

Plot these three points and draw a line through them.

Method 2: Using Gradient and Intercept

1. Plot the y-intercept (0, c).
2. From that point, use the gradient to find another point: go across 1 and up m (or down if m is negative).
3. Draw a line through the points.

Worked Example 4

Plot: y = -3x + 5

Start at (0, 5) — the y-intercept.

Gradient = -3, so from (0, 5) go 1 right and 3 down to reach (1, 2).

Plot (0, 5) and (1, 2) and draw the line through them.

Exam Tip: When plotting a line, always use at least three points. If all three points lie on the same straight line, you can be confident your calculations are correct. If one point is off, you know you need to recheck.

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Finding the Equation from a Graph

If you are given a graph and asked to find the equation:

1. Read the y-intercept (c) from where the line crosses the y-axis.
2. Calculate the gradient (m) by choosing two clear points on the line and using the gradient formula.
3. Write the equation as y = mx + c.

Worked Example 5

A line crosses the y-axis at (0, -1) and passes through (3, 5).

Gradient = (5 - (-1)) / (3 - 0) = 6 / 3 = 2

y-intercept = -1

Equation: y = 2x - 1

Worked Example 6

A line passes through (2, 7) and (6, 3).

Gradient = (3 - 7) / (6 - 2) = -4 / 4 = -1

To find c, substitute one point into y = -x + c:

7 = -(2) + c

c = 9

Equation: y = -x + 9

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Special Lines

Equation	Type	Description
y = c	Horizontal line	Parallel to the x-axis, gradient = 0
x = c	Vertical line	Parallel to the y-axis, gradient undefined
y = x	Diagonal	Passes through origin at 45 degrees, gradient 1
y = -x	Diagonal	Passes through origin at 135 degrees, gradient -1

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Parallel and Perpendicular Lines [H]

Parallel Lines

Two lines are parallel if and only if they have the same gradient.

y = 3x + 1 and y = 3x - 5 are parallel (both have gradient 3).

Perpendicular Lines

Two lines are perpendicular if the product of their gradients is -1. In other words, one gradient is the negative reciprocal of the other.

If line 1 has gradient m, then a perpendicular line has gradient -1/m.

Line 1 gradient	Perpendicular gradient
2	-1/2
-3	1/3
1/4	-4
-2/5	5/2

Worked Example 7

Find the equation of the line perpendicular to y = 2x + 3 that passes through (4, 1).

Gradient of given line = 2

Perpendicular gradient = -1/2

Using y = mx + c with m = -1/2 and the point (4, 1):

1 = (-1/2)(4) + c

1 = -2 + c

c = 3

Equation: y = -x/2 + 3 or equivalently y = -0.5x + 3

Worked Example 8

Show that the lines y = 4x - 1 and x + 4y = 12 are perpendicular.

Rearrange x + 4y = 12: 4y = -x + 12, so y = -x/4 + 3

Gradient of first line = 4

Gradient of second line = -1/4

Product: 4 x (-1/4) = -1

Since the product of the gradients is -1, the lines are perpendicular.

Exam Tip: For perpendicular lines, the gradients multiply to give -1. A quick check: flip the fraction and change the sign. The perpendicular gradient of 3 (which is 3/1) is -1/3.

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Finding the Equation Given a Point and Gradient