Finding a line from information | Year 9

01 / LEARN

The idea, step by step

Use a known point

If the gradient is known, begin with $y = m x + c$ and replace $m$ . Every point on the line must satisfy the equation, so substitute the coordinates of a known point and solve for $c$ . Then write the complete rule using variables again.

Keep $x$ and $y$ in the final answer. They describe all points on the line; the particular point only helped you find the intercept. Substitute that point once more to check your result.

If two points are given

First find the gradient using matching coordinate differences. Then use either point in $y = m x + c$ to determine $c$ . Checking the other point gives an independent test of both the gradient and the intercept.

If the two $x$ -coordinates are equal, the line is vertical: write $x$ equal to that fixed value. If the $y$ -coordinates are equal, it is horizontal: write $y$ equal to that fixed value. Do not divide by zero in the gradient formula.

A common trap

Using the given point's $y$ -coordinate as the intercept.

Instead: The $y$ -intercept is the value when $x = 0$ . Substitute the whole point to calculate it unless the point already lies on the vertical axis.

02 / WORKED EXAMPLES

See the thinking

Example 1

Given a gradient and point

Find the line of gradient 3 through $(2, 7)$ .

Start with $y = 3 x + c$ .
Substitute $(2, 7)$ : $7 = 3 (2) + c$ , so $c = 1$ .
The rule is $y = 3 x + 1$ .
Check: when $x = 2$ , the rule gives $y = 7$ .

y = 3 x + 1

Example 2

Given two points

Find the equation through $(- 1, 4)$ and $(3, - 4)$ .

$m = (- 4 - 4) / (3 - (- 1)) = - 8 / 4 = - 2$ .
Use $(- 1, 4)$ in $y = - 2 x + c$ : $4 = 2 + c$ , so $c = 2$ .
The rule is $y = - 2 x + 2$ .
Check $(3, - 4)$ : $- 2 (3) + 2 = - 4$ .

y = - 2 x + 2

Check your understanding

Find the line of gradient $- 1$ through $(4, 2)$ .

Show answer

$y = - x + 6$

$2 = - 4 + c$ , so $c = 6$ .

03 / WORKSHEET

Your turn

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

Worksheet PDF Worked solutions

Read the questions online

Find the line of gradient 2 through $(1, 5)$ .
2 marks
Find the line of gradient $1 / 2$ through $(4, - 1)$ .
2 marks
Find the equation through $(0, 6)$ and $(3, 0)$ .
3 marks
Find the equation through $(- 2, - 1)$ and $(2, 7)$ .
3 marks
Find the equation through $(5, - 2)$ and $(5, 6)$ .
2 marks
A straight path passes through $(1, 3)$ and $(5, 3)$ . Write its equation and state its gradient.
2 marks

Read the worked solutions

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

Find the line of gradient 2 through $(1, 5)$ .
2 marks
$5 = 2 (1) + c$ gives $c = 3$ .
$y = 2 x + 3$
Find the line of gradient $1 / 2$ through $(4, - 1)$ .
2 marks
$- 1 = 2 + c$ gives $c = - 3$ .
$y = \frac{1}{2} x - 3$
Find the equation through $(0, 6)$ and $(3, 0)$ .
3 marks
$m = (0 - 6) / (3 - 0) = - 2$ .
The first point gives $c = 6$ .
$y = - 2 x + 6$
Find the equation through $(- 2, - 1)$ and $(2, 7)$ .
3 marks
$m = (7 + 1) / (2 + 2) = 2$ .
$- 1 = 2 (- 2) + c$ gives $c = 3$ .
$y = 2 x + 3$
Find the equation through $(5, - 2)$ and $(5, 6)$ .
2 marks
The $x$ -coordinate is 5 at both points.
The line is vertical.
$x = 5$
A straight path passes through $(1, 3)$ and $(5, 3)$ . Write its equation and state its gradient.
2 marks
Both points have the same $y$ -coordinate.
There is no vertical change, so $m = 0$ .
$y = 3$ ; gradient 0.

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Parallel/perpendicular lines as a natural school extension

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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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