Parallel/perpendicular lines as a natural school extension | Year 9

01 / LEARN

The idea, step by step

Same steepness, different line

Two distinct non-vertical straight lines are parallel when they have equal gradients. Their vertical separation stays constant because their $y$ -values change at the same rate. Different intercepts ensure they are not just two equations for the same line.

Rearrange equations before comparing gradients. For example, $2 y = 4 x + 6$ and $y = 2 x - 1$ have gradient 2, so they are parallel; the first line's gradient is not 4.

A right-angle relationship

For two non-vertical, non-horizontal perpendicular lines, their gradients multiply to $- 1$ . Find a perpendicular gradient by taking the reciprocal and changing its sign. Thus a line of gradient $2 / 3$ is perpendicular to one of gradient $- 3 / 2$ .

Horizontal and vertical lines are also perpendicular, but the vertical gradient is undefined, so do not try to apply the multiplication rule to it. Test the equation as well as the gradient when a line must pass through a specified point.

m_{1} m_{2} = - 1

A common trap

Changing only the sign: saying the perpendicular gradient to 3 is $- 3$ .

Instead: The gradients must multiply to $- 1$ . The perpendicular gradient to 3 is $- 1 / 3$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

A parallel line through a point

Find the line parallel to $y = 3 x + 2$ through $(1, - 1)$ .

Parallel lines have the same gradient, so start with $y = 3 x + c$ .
Substitute the point: $- 1 = 3 + c$ , giving $c = - 4$ .
Thus $y = 3 x - 4$ .

y = 3 x - 4

Example 2

A perpendicular line

Find the line perpendicular to $y = \frac{1}{2} x + 4$ through $(2, 1)$ .

The perpendicular gradient is $- 2$ , because $(1 / 2) (- 2) = - 1$ .
Use $y = - 2 x + c$ and substitute $(2, 1)$ : $1 = - 4 + c$ .
Therefore $c = 5$ and $y = - 2 x + 5$ .

y = - 2 x + 5

Check your understanding

What gradient is perpendicular to $- 4$ ?

Show answer

$\frac{1}{4}$

The product $(- 4) (1 / 4) = - 1$ .

03 / WORKSHEET

Your turn

6 questions · 15 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Are $y = 2 x + 5$ and $y = 2 x - 7$ parallel? Explain.
2 marks
Find the gradient of a line perpendicular to $y = - \frac{2}{3} x + 1$ .
2 marks
Find the line parallel to $y = - 3 x + 8$ through $(2, 1)$ .
3 marks
Find the line perpendicular to $y = 2 x - 5$ through $(4, 0)$ .
3 marks
Find a line through $(3, 7)$ perpendicular to $y = 4$ .
2 marks
Show that $3 x + 2 y = 6$ and $2 x - 3 y = 9$ meet at right angles.
3 marks

Read the worked solutions

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

Are $y = 2 x + 5$ and $y = 2 x - 7$ parallel? Explain.
2 marks
Both are in $y = m x + c$ form.
Their gradients are equal and their intercepts differ.
Yes; they are distinct lines with equal gradient 2.
Find the gradient of a line perpendicular to $y = - \frac{2}{3} x + 1$ .
2 marks
Take the negative reciprocal of $- 2 / 3$ .
$(- 2 / 3) (3 / 2) = - 1$ .
$\frac{3}{2}$
Find the line parallel to $y = - 3 x + 8$ through $(2, 1)$ .
3 marks
Use gradient $- 3$ : $1 = - 6 + c$ .
Thus $c = 7$ .
$y = - 3 x + 7$
Find the line perpendicular to $y = 2 x - 5$ through $(4, 0)$ .
3 marks
The perpendicular gradient is $- 1 / 2$ .
$0 = - (1 / 2) (4) + c$ gives $c = 2$ .
$y = - \frac{1}{2} x + 2$
Find a line through $(3, 7)$ perpendicular to $y = 4$ .
2 marks
$y = 4$ is horizontal.
A perpendicular line is vertical and must pass through $x = 3$ .
$x = 3$
Show that $3 x + 2 y = 6$ and $2 x - 3 y = 9$ meet at right angles.
3 marks
Rearrange: $y = - \frac{3}{2} x + 3$ and $y = \frac{2}{3} x - 3$ .
Multiply the gradients: $(- 3 / 2) (2 / 3) = - 1$ .
Their gradients are $- 3 / 2$ and $2 / 3$ , whose product is $- 1$ .

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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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