Straight-Line Graphs

A straight-line graph is the picture of a linear relationship between two variables. Whenever two quantities change at a constant rate — distance against time at a steady speed, total cost against the number of items at a fixed price — the graph is a straight line. This lesson covers plotting straight lines from their equations, finding the gradient and $y$y-intercept, the equation $y = mx + c$y=mx+c, finding the equation of a line from a graph or from given information, and recognising parallel lines and — at Higher tier — perpendicular lines. Reading and drawing graphs is AO1; interpreting gradient and intercept in context is AO3, and explaining the link between an equation and its line is AO2. The whole topic rests on one idea: the equation $y = mx + c$y=mx+c encodes everything about a line in just two numbers, the gradient $m$m and the intercept $c$c.

Key Vocabulary

Term	Meaning
Gradient ($m$m)	The steepness of a line, $\dfrac{\text{change in } y}{\text{change in } x}$change in xchange in y​
$y$y-intercept ($c$c)	The $y$y-value where the line crosses the $y$y-axis
$y = mx + c$y=mx+c	The general equation of a straight line
Coordinate	A pair $(x, y)$(x,y) giving a point's position
Parallel	Lines with the same gradient that never meet
Perpendicular	Lines meeting at $90^{\circ}$90∘; their gradients multiply to $-1$−1
Linear	A relationship whose graph is a straight line

The Equation $y = mx + c$y=mx+c

Every non-vertical straight line can be written as $y = mx + c$y=mx+c, where $m$m is the gradient and $c$c is the $y$y-intercept. Reading these two numbers straight from the equation lets you sketch the line quickly.

The graph below shows the line $y = 2x - 1$y=2x−1. Its gradient is $2$2 (it rises $2$2 for every $1$1 across) and its $y$y-intercept is $-1$−1 (it crosses the $y$y-axis at $-1$−1).

Worked Example 1

Write down the gradient and $y$y-intercept of $y = 3x + 5$y=3x+5.

The equation is in the form $y = mx + c$y=mx+c, so $m = 3$m=3 and $c = 5$c=5.

Answer: gradient $3$3, $y$y-intercept $5$5.

Worked Example 2

Rearrange $2y = 6x - 8$2y=6x−8 into the form $y = mx + c$y=mx+c and write down its gradient.

Divide every term by $2$2: $y = 3x - 4$y=3x−4. So the gradient is $3$3.

Answer: $y = 3x - 4$y=3x−4; gradient $3$3.

Common error: reading the gradient as $6$6 straight from $2y = 6x - 8$2y=6x−8 without first making $y$y the subject.

Worked Example 2b

Write down the gradient and $y$y-intercept of $y = 7 - 2x$y=7−2x.

Rearrange mentally into the form $y = mx + c$y=mx+c: $y = -2x + 7$y=−2x+7. So the gradient is $-2$−2 and the $y$y-intercept is $7$7.

Answer: gradient $-2$−2, $y$y-intercept $7$7.

Common error: reading the gradient as $7$7 because it is written first. The gradient is always the number multiplying $x$x, regardless of the order the terms appear in.

Finding the Gradient

The gradient measures steepness as $\dfrac{\text{change in } y}{\text{change in } x}$change in xchange in y​ between any two points on the line. A line going uphill from left to right has a positive gradient; one going downhill has a negative gradient; a horizontal line has gradient $0$0. The steeper the line, the larger the size of the gradient. A useful phrase is "rise over run": the rise is how far up you go, the run is how far across.

Worked Example 3

Work out the gradient of the line through $(1, 2)$(1,2) and $(4, 11)$(4,11).

Gradient $= \dfrac{11 - 2}{4 - 1} = \dfrac{9}{3} = 3$=4−111−2​=39​=3.

Answer: $3$3

Worked Example 4

Work out the gradient of the line through $(-2, 7)$(−2,7) and $(2, -1)$(2,−1).

Gradient $= \dfrac{-1 - 7}{2 - (-2)} = \dfrac{-8}{4} = -2$=2−(−2)−1−7​=4−8​=−2.

Answer: $-2$−2

Common error: subtracting the coordinates in a different order on the top and bottom; keep the same point first in both.

Finding the Equation of a Line

Given the gradient and one point, substitute into $y = mx + c$y=mx+c to find $c$c. Given two points, find the gradient first, then do the same. The logic is always the same: once you know the gradient $m$m, the only missing piece is $c$c, and substituting any point the line passes through gives you one equation to solve for it. There is no need to use both points to find $c$c — one is enough — though substituting the second point at the end is an excellent way to check your answer.

Worked Example 5

A line has gradient $4$4 and passes through $(2, 11)$(2,11). Find its equation.

Use $y = 4x + c$y=4x+c. Substitute $(2, 11)$(2,11): $11 = 4(2) + c$11=4(2)+c, so $11 = 8 + c$11=8+c and $c = 3$c=3.

Answer: $y = 4x + 3$y=4x+3

Worked Example 6

Find the equation of the line through $(1, 5)$(1,5) and $(3, 11)$(3,11).

Gradient $= \dfrac{11 - 5}{3 - 1} = \dfrac{6}{2} = 3$=3−111−5​=26​=3. Use $y = 3x + c$y=3x+c with $(1, 5)$(1,5): $5 = 3(1) + c$5=3(1)+c, so $c = 2$c=2.

Answer: $y = 3x + 2$y=3x+2

Worked Example 6b

Find the equation of the line through $(-2, 1)$(−2,1) and $(2, 9)$(2,9).

Gradient $= \dfrac{9 - 1}{2 - (-2)} = \dfrac{8}{4} = 2$=2−(−2)9−1​=48​=2. Use $y = 2x + c$y=2x+c with the point $(2, 9)$(2,9): $9 = 2(2) + c$9=2(2)+c, so $9 = 4 + c$9=4+c and $c = 5$c=5.

Answer: $y = 2x + 5$y=2x+5

A small but important detail is the subtraction in the denominator: $2 - (-2) = 4$2−(−2)=4, not $0$0. Mishandling the double negative when one coordinate is negative is a frequent source of error in gradient calculations.

Parallel and Perpendicular Lines

Parallel lines have the same gradient, which is why they never meet — they rise at exactly the same rate. Perpendicular lines [Higher] meet at a right angle, and their gradients multiply to $-1$−1 — so one gradient is the negative reciprocal of the other. To find a negative reciprocal, turn the fraction upside down and change its sign: the negative reciprocal of $3$3 (i.e. $\dfrac{3}{1}$13​) is $-\dfrac{1}{3}$−31​, and the negative reciprocal of $-\dfrac{2}{5}$−52​ is $\dfrac{5}{2}$25​.

Worked Example 7

Write down the equation of the line parallel to $y = 5x - 2$y=5x−2 that passes through $(0, 3)$(0,3).

Parallel means the same gradient, $5$5, and the $y$y-intercept here is $3$3.

Answer: $y = 5x + 3$y=5x+3

Worked Example 8 — perpendicular [Higher]