Gradient as rise/run | Year 9

01 / LEARN

Understand the idea.

Rise divided by run

Gradient measures the vertical change for each unit of horizontal change. Choose two distinct points, label them in an order, then subtract the coordinates in that same order in both parts of the fraction.

The ratio stays constant along a straight line. Reading from left to right, rising lines have positive gradient, falling lines have negative gradient, and horizontal lines have gradient zero. A vertical line has zero run, so its gradient is undefined.

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

A solid segment joins the labelled points A at negative 2,1 and B at 4,4. A dashed horizontal-then-vertical construction shows the matching run and rise on an equal-scale grid.

Read the direction and units

A gradient of $\frac{3}{2}$ means a rise of $3$ for a run of $2$ , or a rise of $1.5$ for a run of $1$ . It describes a change, rather than the line's starting height.

Attach the graph's units to the ratio. If vertical coordinates are dollars and horizontal coordinates are hours, gradient is measured in dollars per hour. A negative gradient then describes a decrease in cost as time increases.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A positive gradient

Find the gradient through $A (- 2, 1)$ and $B (4, 4)$ .

Worked solution

1
Find the vertical change.
$Δ y = 4 - 1 = 3$
Read both coordinates from A to B.
2
Find the matching horizontal change.
$Δ x = 4 - (- 2) = 6$
Subtract the starting x-coordinate, including its negative sign.
3
Divide rise by run.
$m = \frac{3}{6} = \frac{1}{2}$
The line rises one unit for every two units to the right.

Answer

$m = \frac{1}{2}$

EXAMPLE 02

A falling line

Find the gradient through $(1, 7)$ and $(5, - 1)$ .

Worked solution

1
Subtract in the same order.
$\begin{aligned} Δ y & = - 1 - 7 = - 8 \\ Δ x & = 5 - 1 = 4 \end{aligned}$
Both changes run from the first point to the second.
2
Calculate the gradient.
$m = \frac{- 8}{4} = - 2$
Gradient is vertical change divided by horizontal change.
3
Interpret the sign and magnitude.
$Δ x = 1 \Rightarrow Δ y = - 2$
The line falls two units for each unit moved to the right.

Answer

$m = - 2$

03 / YOUR TURN

Try it yourself.

Check your understanding

What is the gradient through $(0, 5)$ and $(4, 5)$ ?

Show answer & reasoning

The coordinates have the same height: $m = \frac{5 - 5}{4 - 0} = \frac{0}{4} = 0$ . This is a horizontal line, with no vertical change.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Find the gradient through $(1, 2)$ and $(4, 8)$ .
2 marks
Find the gradient through $(- 3, 5)$ and $(1, - 3)$ .
2 marks
Find the gradient through $(2, - 1)$ and $(6, 1)$ .
2 marks
Describe the gradient of the line $x = 4$ .
2 marks
A tank contains 80 L at 2 minutes and 50 L at 8 minutes. Find the average gradient and explain it.
3 marks
Two points on a line are $(- 2, 3)$ and $(4, k)$ . Its gradient is $\frac{2}{3}$ . Find $k$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the gradient through $(1, 2)$ and $(4, 8)$ .
2 marks
$m = \frac{8 - 2}{4 - 1} = \frac{6}{3} = 2$ .
2
Find the gradient through $(- 3, 5)$ and $(1, - 3)$ .
2 marks
$m = \frac{- 3 - 5}{1 - (- 3)} = - \frac{8}{4} = - 2$ .
$- 2$
Find the gradient through $(2, - 1)$ and $(6, 1)$ .
2 marks
$m = \frac{1 + 1}{6 - 2} = \frac{2}{4} = \frac{1}{2}$ .
$\frac{1}{2}$
Describe the gradient of the line $x = 4$ .
2 marks
Every point has the same $x$ -coordinate.
The run is zero, so division by the run is not defined.
Undefined
A tank contains 80 L at 2 minutes and 50 L at 8 minutes. Find the average gradient and explain it.
3 marks
$m = \frac{50 - 80}{8 - 2} = - \frac{30}{6} = - 5$ .
The negative sign represents a decrease.
$- 5$ L/min; the tank loses an average of 5 L per minute over this interval.
Two points on a line are $(- 2, 3)$ and $(4, k)$ . Its gradient is $\frac{2}{3}$ . Find $k$ .
3 marks
$\frac{k - 3}{4 + 2} = \frac{2}{3}$ .
Thus $k - 3 = 4$ , giving $k = 7$ .
$k = 7$

Checking saved progress…

Curriculum reference & lesson scope

Australian Curriculum v9 and SA Curriculum Mathematics

Gradient, midpoint and distance; algebraic and graphical straight-line relationships; interpreting models and testing parameters. Parallel and perpendicular lines are included as a natural school-course extension. All explanations and questions are original.

Official curriculum source ↗

In this lesson

Understand the idea.

Rise divided by run

Read the direction and units

See the method in action.

A positive gradient

Find the vertical change.

Find the matching horizontal change.

Divide rise by run.

A falling line

Subtract in the same order.

Calculate the gradient.

Interpret the sign and magnitude.

Try it yourself.

Check your understanding

Keep practising