Straight-line equations and intercepts | Year 9

01 / LEARN

Understand the idea.

What $m$ and $c$ tell you

In gradient-intercept form, $y = m x + c$ , the coefficient $m$ gives the rise per unit run. When $x = 0$ , the product $m x$ vanishes, leaving $y = c$ ; this locates the point $(0, c)$ on the vertical axis.

Rearrange before reading the gradient. In $2 y = 6 x + 8$ , dividing every term by $2$ gives $y = 3 x + 4$ . The gradient is therefore $3$ . From the intercept, use that gradient to place a second point.

y = m x + c

Equal-unit Cartesian grid from negative six to eight on both axes, showing y equals two x minus two. The intercepts zero, negative two and one, zero are marked and labelled.

Find intercepts deliberately

An intercept lies on an axis. On the horizontal axis, $y = 0$ , so set $y$ to zero to find an $x$ -intercept. On the vertical axis, $x = 0$ , so set $x$ to zero instead.

Write each intercept as an ordered pair to show which coordinate is zero. Horizontal lines have form $y = k$ and gradient zero. Vertical lines have form $x = k$ and require no numerical gradient or gradient-intercept form.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Read a rule

For $y = - 3 x + 6$ , state the gradient and both axis intercepts.

Worked solution

1
Match the gradient-intercept form.
$y = - 3 x + 6 \Rightarrow m = - 3, c = 6$
The equation already has y isolated with coefficient 1.
2
Find the vertical-axis crossing.
$x = 0 \Rightarrow y = 6, (0, 6)$
Every point on the y-axis has x-coordinate zero.
3
Set the height to zero.
$\begin{aligned} 0 & = - 3 x + 6 \\ 3 x & = 6 \\ x & = 2 \end{aligned}$
The x-axis consists of points whose y-coordinate is zero.
4
State the horizontal-axis crossing.
$(2, 0)$
The solved x-value pairs with y equal to zero.

Answer

Gradient $- 3$ ; intercepts $(0, 6)$ and $(2, 0)$ .

EXAMPLE 02

Rearrange first

Find the gradient and $y$ -intercept of $3 x + 2 y = 8$ .

Worked solution

1
Move the x-term to the other side.
$2 y = - 3 x + 8$
Subtracting 3x from both sides preserves equality.
2
Divide every term by two.
$y = - \frac{3}{2} x + 4$
The coefficient of y must be 1 before reading the gradient.
3
Read the required features.
$m = - \frac{3}{2}, (0, 4)$
The coefficient of x is the gradient; the constant is the y-value at x equal to zero.

Answer

Gradient $- \frac{3}{2}$ ; $y$ -intercept $(0, 4)$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

What are the gradient and $y$ -intercept of $y = \frac{1}{2} x - 4$ ?

Show answer & reasoning

Gradient $\frac{1}{2}$ ; $y$ -intercept $(0, - 4)$ .

Comparing $y = \frac{1}{2} x - 4$ with $y = m x + c$ gives $m = \frac{1}{2}$ and $c = - 4$ . The intercept is the point $(0, - 4)$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

State the gradient and $y$ -intercept of $y = 4 x - 7$ .
2 marks
Find both intercepts of $y = - 2 x + 8$ .
3 marks
Write $4 x + 2 y = 10$ in gradient–intercept form.
2 marks
Give three points on $y = \frac{3}{2} x + 1$ using $x = 0, 2, 4$ .
3 marks
Does the point $(3, - 1)$ lie on $y = 2 x - 7$ ? Explain.
2 marks
Describe and sketch the line $x = - 2$ .
2 marks

Read the worked solutions

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

State the gradient and $y$ -intercept of $y = 4 x - 7$ .
2 marks
Match $m = 4$ and $c = - 7$ .
Gradient 4; $(0, - 7)$ .
Find both intercepts of $y = - 2 x + 8$ .
3 marks
Set $x = 0$ to get $y = 8$ .
Set $y = 0$ : $2 x = 8$ , so $x = 4$ .
$(0, 8)$ and $(4, 0)$ .
Write $4 x + 2 y = 10$ in gradient–intercept form.
2 marks
$2 y = - 4 x + 10$ .
Divide by 2.
$y = - 2 x + 5$
Give three points on $y = \frac{3}{2} x + 1$ using $x = 0, 2, 4$ .
3 marks
Substitute each value into $y = 1.5 x + 1$ .
The resulting values are 1, 4 and 7.
$(0, 1)$ , $(2, 4)$ , $(4, 7)$ .
Does the point $(3, - 1)$ lie on $y = 2 x - 7$ ? Explain.
2 marks
Substitute $x = 3$ : $y = 6 - 7 = - 1$ .
This matches the given $y$ -coordinate.
Yes.
Describe and sketch the line $x = - 2$ .
2 marks
Two points are $(- 2, 0)$ and $(- 2, 3)$ .
Join them vertically; its gradient is undefined.
A vertical line through all points whose $x$ -coordinate is $- 2$ .

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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 ↗

In this lesson

Understand the idea.

What $m$ and $c$ tell you

Find intercepts deliberately

See the method in action.

Read a rule

Match the gradient-intercept form.

Find the vertical-axis crossing.

Set the height to zero.

State the horizontal-axis crossing.

Rearrange first

Move the x-term to the other side.

Divide every term by two.

Read the required features.

Try it yourself.

Check your understanding

Keep practising