Finding a line from points or gradient | Year 11

01 / LEARN

Understand the idea.

Start from a known point

A non-vertical line is determined by its gradient $m$ and one point $(x_{1}, y_{1})$ . The equation $y - y_{1} = m (x - x_{1})$ states that every change from the known point has the required rise-to-run ratio.

Insert signed coordinates carefully: subtracting $x_{1} = - 2$ creates $x + 2$ . Point-gradient form is already a complete equation of the line. Expand only when another form is useful or requested.

y - y_{1} = m (x - x_{1})

Using two points or intercepts

From two distinct points, calculate the gradient and then use either point to find the line. Substituting the other point checks the result independently. If their horizontal coordinates agree, write a vertical line $x = constant$ instead of dividing by zero.

Intercepts are simply particular points: a horizontal-axis intercept $a$ supplies $(a, 0)$ , while a vertical-axis intercept $b$ supplies $(0, b)$ . In a model, give meaningful input restrictions as well as an equation, so a mathematical extension is not mistaken for a practical prediction.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A gradient and a point

Find the line of gradient 3 through $(- 2, 4)$ .

Worked solution

1
Substitute the known point and gradient.
$y - 4 = 3 (x - (- 2)) = 3 (x + 2)$
The differences are measured from the point with coordinates negative two and four.
2
Expand and isolate the output.
$\begin{aligned} y - 4 & = 3 x + 6 \\ y & = 3 x + 10 \end{aligned}$
Distributing three and adding four preserve the equation.
3
Check the given point.
$3 (- 2) + 10 = 4$
The rule gives the required output, and its coefficient of x is the required gradient three.

Answer

$y = 3 x + 10$

EXAMPLE 02

Two points

Find the equation through (1,3) and (4,9).

Worked solution

1
Calculate the gradient.
$m = \frac{9 - 3}{4 - 1} = \frac{6}{3} = 2$
The point order is the same in both differences.
2
Use one point to find the intercept.
$\begin{aligned} 3 & = 2 (1) + c \\ c & = 1 \end{aligned}$
The point with input one must satisfy the equation of the line.
3
Write and check the finished rule.
$y = 2 x + 1, 2 (4) + 1 = 9$
The second point also satisfies the equation, confirming the line through both points.

Answer

$y = 2 x + 1$

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the line of gradient -2 through (3,1).

Show answer & reasoning

$y = - 2 x + 7$ .

Use $y - 1 = - 2 (x - 3)$ . Expanding gives $y = - 2 x + 7$ , and substituting $x = 3$ returns $y = 1$ .

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Find the line of gradient 4 through $(2, 1)$ .
2 marks
Find the line through $(- 1, 2)$ and $(3, 10)$ .
2 marks
A line meets the axes at $(6, 0)$ and $(0, 3)$ . Find its equation.
2 marks
Find the equation through $(- 4, 2)$ and $(- 4, 9)$ .
2 marks
A taxi fare is linear. A 2 km trip costs 11 dollars and a 7 km trip costs 23.50 dollars. Find a fare model.
2 marks
Does $y = 3 x - 1$ pass through both $(2, 5)$ and $(4, 10)$ ? Explain.
2 marks

Read the worked solutions

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

Find the line of gradient 4 through $(2, 1)$ .
2 marks
$y - 1 = 4 (x - 2)$ .
$y = 4 x - 7$
Find the line through $(- 1, 2)$ and $(3, 10)$ .
2 marks
$Gradient = \frac{10 - 2}{3 + 1} = 2$ .
Use $(- 1, 2)$ : $2 = - 2 + c$ , so $c = 4$ .
$y = 2 x + 4$
A line meets the axes at $(6, 0)$ and $(0, 3)$ . Find its equation.
2 marks
$Gradient = \frac{0 - 3}{6 - 0} = \frac{- 1}{2}$ and the y-intercept is 3.
$y = \frac{- x}{2} + 3$
Find the equation through $(- 4, 2)$ and $(- 4, 9)$ .
2 marks
Both x-coordinates agree, so the line is vertical.
$x = - 4$
A taxi fare is linear. A 2 km trip costs 11 dollars and a 7 km trip costs 23.50 dollars. Find a fare model.
2 marks
$Rate = \frac{23.50 - 11}{7 - 2} = 2.50$ dollars/km.
Using $d = 2$ gives the initial fee 6 dollars.
$C (d) = 2.5 d + 6$ for distances $d$ in kilometres.
Does $y = 3 x - 1$ pass through both $(2, 5)$ and $(4, 10)$ ? Explain.
2 marks
At $x = 2$ , $y = 5$ ; at $x = 4$ , the rule gives 11 rather than 10.
It passes through $(2, 5)$ only.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics, Topic 1: Functions and graphs. Prerequisite algebra and coordinate geometry are embedded as supporting lessons.

Official curriculum source ↗

In this lesson

Understand the idea.

Start from a known point

Using two points or intercepts

See the method in action.

A gradient and a point

Substitute the known point and gradient.

Expand and isolate the output.

Check the given point.

Two points

Calculate the gradient.

Use one point to find the intercept.

Write and check the finished rule.

Try it yourself.

Check your understanding

Keep practising