Straight-line equations and intercepts | Year 9

01 / LEARN

The idea, step by step

What $m$ and $c$ tell you

In the form $y = m x + c$ , $m$ is the gradient and $c$ is the value of $y$ when $x = 0$ . The point $(0, c)$ is the vertical, or $y$ -, intercept. Starting there, use rise/run to locate a second point and draw the straight line through both.

The coefficient of $x$ is not always the gradient until the equation has been rearranged into this form. For example, $2 y = 6 x + 8$ becomes $y = 3 x + 4$ , so its gradient is 3, not 6.

y = m x + c

Find intercepts deliberately

Every point on the horizontal axis has $y = 0$ , so substitute $y = 0$ to find an $x$ -intercept. Every point on the vertical axis has $x = 0$ , so substitute $x = 0$ for the $y$ -intercept. Write intercepts as points, not as a pair of unrelated numbers.

A horizontal line has the form $y = k$ and gradient zero. A vertical line has the form $x = k$ ; it cannot be written as $y = m x + c$ with an ordinary numerical gradient. Always check whether a claimed intercept really satisfies the equation.

A common trap

Setting $x = 0$ to find the $x$ -intercept.

Instead: At an $x$ -intercept, the point lies on the horizontal axis, so its $y$ -coordinate is zero.

02 / WORKED EXAMPLES

See the thinking

Example 1

Read a rule

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

The equation is already $y = m x + c$ , so $m = - 3$ and the $y$ -intercept is $(0, 6)$ .
Set $y = 0$ : $0 = - 3 x + 6$ , giving $x = 2$ .
Thus the $x$ -intercept is $(2, 0)$ .

Gradient

- 3

; intercepts

(0, 6)

and

(2, 0)

Example 2

Rearrange first

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

Subtract $3 x$ : $2 y = - 3 x + 8$ .
Divide every term by 2: $y = - \frac{3}{2} x + 4$ .
Read $m = - 3 / 2$ and $c = 4$ .

Gradient

- \frac{3}{2}

;

y

-intercept

(0, 4)

Check your understanding

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

Show answer

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

Match the equation to $y = m x + c$ : $m = 1 / 2$ and $c = - 4$ .

03 / WORKSHEET

Your turn

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

Worksheet PDF 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 ↗

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