Gradient as rise/run | Year 9

01 / LEARN

The idea, step by step

Rise divided by run

Choose two different points on a straight line. The rise is the change in the $y$ -coordinate and the run is the matching change in the $x$ -coordinate. Use the same direction for both changes; do not subtract one pair forwards and the other backwards.

A straight line has the same gradient everywhere. A rising line has positive gradient, a falling line has negative gradient, and a horizontal line has gradient zero. A vertical line has zero horizontal change, so its gradient is undefined.

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

Read the direction and units

A gradient of $3 / 2$ means $y$ increases by 3 when $x$ increases by 2. It does not mean the line must start at the origin. Any two points on that line give the same ratio.

In a context, gradient has units of vertical quantity per horizontal quantity. For a cost–time graph, it might mean dollars per hour. Always say what a rise or fall means for the situation.

A common trap

Using horizontal change divided by vertical change.

Instead: Gradient is vertical change divided by horizontal change. Sketch a small right triangle and label rise and run first.

02 / WORKED EXAMPLES

See the thinking

Example 1

A positive gradient

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

The vertical change is $4 - 1 = 3$ .
The horizontal change is $4 - (- 2) = 6$ .
Therefore $m = 3 / 6 = 1 / 2$ .

m = \frac{1}{2}

Example 2

A falling line

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

Use matching order: $m = (- 1 - 7) / (5 - 1)$ .
This is $- 8 / 4 = - 2$ .
The line falls 2 units for each unit to the right.

m = - 2

Check your understanding

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

Show answer

The rise is $5 - 5 = 0$ and the run is 4, so the line is horizontal.

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF 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 $2 / 3$ . 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
$m = (8 - 2) / (4 - 1) = 6 / 3 = 2$ .
2
Find the gradient through $(- 3, 5)$ and $(1, - 3)$ .
2 marks
$m = (- 3 - 5) / (1 - (- 3)) = - 8 / 4 = - 2$ .
$- 2$
Find the gradient through $(2, - 1)$ and $(6, 1)$ .
2 marks
$m = (1 + 1) / (6 - 2) = 2 / 4 = 1 / 2$ .
$\frac{1}{2}$
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
$m = (50 - 80) / (8 - 2) = - 30 / 6 = - 5$ .
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 $2 / 3$ . Find $k$ .
3 marks
$(k - 3) / (4 + 2) = 2 / 3$ .
Thus $k - 3 = 4$ , giving $k = 7$ .
$k = 7$

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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 ↗

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