Finding a line from points or gradient | Year 11

01 / LEARN

The idea, step by step

Start from a known point

If a non-vertical line has gradient m and passes through $(x_{1}, y_{1})$ , any point on it has vertical change $y - y_{1}$ and horizontal change $x - x_{1}$ . Equating their ratio to m gives the point-gradient form. It is particularly useful when the known point is not an intercept.

Insert the point with its original signs. If $x_{1} = - 2$ , the bracket is $x + 2$ . You can leave a correct equation in point-gradient form unless another form is requested; expanding too early often introduces avoidable errors.

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

Using two points or intercepts

With two distinct points, calculate the gradient first, then use either point to find the remaining constant. Check the finished rule with the other point. If the points have the same x-coordinate, the line is vertical and should be written x=constant rather than forcing a slope-intercept form.

When the two intercepts are supplied, treat them as coordinate pairs. In a context, explain the input restrictions as well as the equation: a valid straight line can still predict impossible values outside the situation being modelled.

A common trap

Using the y-coordinate of a known point as the intercept even when its x-coordinate is not zero.

Instead: Substitute the full point into $y = m x + c$ to calculate c. Only a point with x=0 directly supplies the y-intercept.

02 / WORKED EXAMPLES

See the thinking

Example 1

A gradient and a point

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

Use $y - 4 = 3 (x + 2)$ .
Expand and add 4: y=3x+10. Check: $3 (- 2) + 10 = 4$ .

y = 3 x + 10

Example 2

Two points

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

The gradient is $(9 - 3) / (4 - 1) = 2$ .
Use $3 = 2 (1) + c$ , so c=1. Check at x=4: y=9.

y = 2 x + 1

Check your understanding

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

Show answer

$y = - 2 x + 7$ .

$y - 1 = - 2 (x - 3)$ gives the rule; substitution checks the point.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF 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=(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=(0-3)/(6-0)=-1/2 and the y-intercept is 3.
$y = - x / 2 + 3$
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=(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 ↗

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