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

01 / LEARN

Understand the idea.

Same steepness, different line

Distinct non-vertical parallel lines have equal gradients. At matching horizontal positions, their outputs differ by a constant amount. Different intercepts distinguish two parallel lines from two equations describing the same line.

Isolate $y$ before comparing gradients. For example, $2 y = 4 x + 6$ becomes $y = 2 x + 3$ , so it is parallel to $y = 2 x - 1$ . Distinct vertical lines are parallel too, although their gradients are undefined.

Equal-unit Cartesian grid from negative six to eight on both axes, showing the labelled parallel lines y equals x plus two in gold and y equals x minus two in sage.

A right-angle relationship

For perpendicular lines with finite non-zero gradients, $m_{1} m_{2} = - 1$ . Taking the reciprocal and changing the sign produces the required partner. Thus $\frac{2}{3}$ pairs with $- \frac{3}{2}$ , since their product is $- 1$ .

Horizontal and vertical lines also meet at right angles, but the gradient product cannot describe that case. When a required line passes through a specified point, determine its intercept as well as its gradient.

m_{1} m_{2} = - 1

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A parallel line through a point

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

Worked solution

1
Use the same gradient.
$m = 3, y = 3 x + c$
A parallel non-vertical line has the original line's gradient.
2
Substitute the required point.
$- 1 = 3 (1) + c$
The line must pass through (1,-1).
3
Solve for its intercept.
$c = - 1 - 3 = - 4$
Subtract the x-term from the point's y-coordinate.
4
Write and verify the line.
$y = 3 x - 4, 3 (1) - 4 = - 1$
Its gradient is 3 and its different intercept makes it a distinct parallel line.

Answer

$y = 3 x - 4$

EXAMPLE 02

A perpendicular line

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

Worked solution

1
Find the perpendicular gradient.
$m = - \frac{1}{\frac{1}{2}} = - 2, \frac{1}{2} (- 2) = - 1$
Finite non-zero perpendicular gradients multiply to -1.
2
Use the specified point.
$1 = - 2 (2) + c = - 4 + c$
Substitution places the new line through (2,1).
3
Find the intercept and rule.
$c = 5, y = - 2 x + 5$
Adding 4 to both sides isolates the intercept.
4
Check the point.
$- 2 (2) + 5 = 1$
The equation satisfies both the required gradient and location.

Answer

$y = - 2 x + 5$

03 / YOUR TURN

Try it yourself.

Check your understanding

What gradient is perpendicular to $- 4$ ?

Show answer & reasoning

$\frac{1}{4}$

The negative reciprocal of $- 4$ is $\frac{1}{4}$ . Check the defining relationship: $(- 4) (\frac{1}{4}) = - 1$ .

Keep practising

6 worksheet questions · 15 marks

Open worksheet 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 $- \frac{2}{3}$ .
$(- \frac{2}{3}) (\frac{3}{2}) = - 1$ .
$\frac{3}{2}$
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 $- \frac{1}{2}$ .
$0 = - (\frac{1}{2}) (4) + c$ 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: $(- \frac{3}{2}) (\frac{2}{3}) = - 1$ .
Their gradients are $- \frac{3}{2}$ and $\frac{2}{3}$ , 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 ↗

In this lesson

Understand the idea.

Same steepness, different line

A right-angle relationship

See the method in action.

A parallel line through a point

Use the same gradient.

Substitute the required point.

Solve for its intercept.

Write and verify the line.

A perpendicular line

Find the perpendicular gradient.

Use the specified point.

Find the intercept and rule.

Check the point.

Try it yourself.

Check your understanding

Keep practising