Finding a line from information | Year 9

01 / LEARN

Understand the idea.

Use a known point

A known gradient determines the steepness, but many parallel lines share it. Substitute that gradient into $y = m x + c$ , then use the given point to determine which intercept places the line correctly.

Replace both coordinates together: a point $(a, b)$ means $x = a$ and $y = b$ . Solve the resulting equation for $c$ , then return to variables in the final rule so it describes every point on the line.

If two points are given

With two distinct points, first calculate the gradient using matching coordinate differences. Use either point to find the intercept, then substitute the other point as a separate check of the complete equation.

Check for special cases before dividing. Equal $x$ -coordinates give a vertical line $x = k$ with undefined gradient. Equal $y$ -coordinates give a horizontal line $y = k$ with gradient zero. Both still have valid equations.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Given a gradient and point

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

Worked solution

1
Insert the known gradient.
$y = 3 x + c$
Every non-vertical line of gradient 3 has this form.
2
Use the given point.
$7 = 3 (2) + c$
The coordinates of every point on the line satisfy its equation.
3
Solve for the intercept.
$c = 7 - 6 = 1$
Subtract the contribution from the x-term.
4
Write and check the full rule.
$y = 3 x + 1, 3 (2) + 1 = 7$
The rule has gradient 3 and passes through the required point.

Answer

$y = 3 x + 1$

EXAMPLE 02

Given two points

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

Equal-unit Cartesian grid from negative six to eight on both axes. The gold line y equals negative two x plus two passes through the marked points A at negative one, four and B at three, negative four. The original equation and letter labels are retained.

Worked solution

1
Calculate the gradient.
$m = \frac{- 4 - 4}{3 - (- 1)} = \frac{- 8}{4} = - 2$
Use the same direction for both coordinate changes.
2
Substitute the first point.
$4 = - 2 (- 1) + c = 2 + c$
The point (-1,4) lies on the required line.
3
Find the intercept and equation.
$c = 2, y = - 2 x + 2$
Subtracting 2 isolates the unknown intercept.
4
Check the second point.
$- 2 (3) + 2 = - 4$
The other endpoint independently confirms that the line passes through both points.

Answer

$y = - 2 x + 2$

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$y = - x + 6$

Begin with $y = - x + c$ . Substituting $(4, 2)$ gives $2 = - 4 + c$ , hence $c = 6$ and $y = - x + 6$ . The point checks because $- 4 + 6 = 2$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Find the line of gradient 2 through $(1, 5)$ .
2 marks
Find the line of gradient $\frac{1}{2}$ 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 $\frac{1}{2}$ 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
$m = \frac{0 - 6}{3 - 0} = - 2$ .
The first point gives $c = 6$ .
$y = - 2 x + 6$
Find the equation through $(- 2, - 1)$ and $(2, 7)$ .
3 marks
$m = \frac{7 + 1}{2 + 2} = 2$ .
$- 1 = 2 (- 2) + c$ 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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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.

Use a known point

If two points are given

See the method in action.

Given a gradient and point

Insert the known gradient.

Use the given point.

Solve for the intercept.

Write and check the full rule.

Given two points

Calculate the gradient.

Substitute the first point.

Find the intercept and equation.

Check the second point.

Try it yourself.

Check your understanding

Keep practising