Calculating the gradient of a straight line from two points using the gradient formula, and interpreting positive, negative, zero and undefined gradients.

What this dot point is asking

The SQA wants you to calculate the gradient of a straight line from the coordinates of two points using the gradient formula, and to interpret what the sign and size of a gradient mean: how steep the line is and whether it rises, falls, is level or vertical.

The gradient formula

The gradient between two points $(x_1, y_1)$ and $(x_2, y_2)$ is the vertical change divided by the horizontal change between them.

Reading the sign of a gradient

The sign of the gradient tells you the direction of the slope as you read the line from left to right.

For example, the line through $(0, 4)$ and $(3, 1)$ has gradient $\dfrac{1 - 4}{3 - 0} = \dfrac{-3}{3} = -1$, so it falls; the line through $(2, 5)$ and $(6, 5)$ has gradient $\dfrac{5 - 5}{6 - 2} = 0$, so it is horizontal.

Steepness and size

The further the gradient is from zero, the steeper the line. A gradient of $4$ is steeper than a gradient of $2$, and a gradient of $-3$ is steeper than a gradient of $-1$. This is why gradient is used for road and ramp design: a road described as "1 in 5" has gradient $\dfrac{1}{5} = 0.2$.

When comparing a positive gradient with a negative one, compare their sizes ignoring the sign to judge steepness, but use the sign to judge direction. A line of gradient $-4$ is steeper than one of gradient $2$, even though $-4$ is the smaller number, because its size $4$ is larger.

Working backwards from a gradient

Many questions give you the gradient and one point, then ask for the missing coordinate of a second point. You rearrange the gradient formula to do this. For a line of gradient $m$ through $(x_1, y_1)$ and $(x_2, y_2)$, the relationship $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ can be rearranged to $y_2 - y_1 = m(x_2 - x_1)$, which lets you find any missing value.

Examples in context

Gradient describes any rate of steepness or change. A wheelchair ramp must not be too steep, so building regulations limit its gradient, often to about $\dfrac{1}{12}$, meaning a rise of $1$ for every $12$ of horizontal run. On a distance-time graph the gradient is the speed, so a steeper line means faster motion; a horizontal section (zero gradient) means the object is stationary. The same formula underpins the equation of a straight line that follows in the Relationships area.

Try this

Q1. Find the gradient of the line through $(2, 3)$ and $(8, 15)$. [2 marks]

• Cue. $\dfrac{15 - 3}{8 - 2} = \dfrac{12}{6} = 2$.

Q2. Find the gradient of the line through $(-1, 5)$ and $(3, -3)$. [2 marks]

• Cue. $\dfrac{-3 - 5}{3 - (-1)} = \dfrac{-8}{4} = -2$.

Q3. State the gradient of a horizontal line. [1 mark]

• Cue. $0$.