Parallel and perpendicular lines | Year 11

01 / LEARN

Understand the idea.

Parallel lines

Distinct non-vertical lines are parallel when their gradients match. If their intercepts also match, they are the same line. Rearranging each equation into $y = m x + c$ separates these two possibilities.

Vertical lines provide an important exception to calculations with gradients. Distinct equations $x = a$ and $x = b$ are parallel even though neither has a defined finite gradient. Identify a vertical line before applying slope formulas.

Perpendicular lines

With equal geometric units on both axes, perpendicular lines with finite nonzero gradients satisfy $m_{1} m_{2} = - 1$ . Take the negative reciprocal of the original gradient: both its sign and its numerator-denominator relationship change.

A horizontal line is perpendicular to a vertical line. When the original gradient is zero, write the perpendicular line as $x = constant$ through the required point. For other gradients, use point-gradient form after finding the negative reciprocal.

m_{1} m_{2} = - 1

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A parallel line

Find the line parallel to $2 x - y = 7$ through (3,4).

Worked solution

1
Read the original gradient after rearranging.
$2 x - y = 7 ⟹ y = 2 x - 7$
The coefficient of x is two in slope-intercept form.
2
Use the same gradient through the required point.
$y - 4 = 2 (x - 3)$
A parallel line retains the gradient and must pass through the supplied point.
3
Simplify and distinguish the lines.
$y = 2 x - 2$
Its gradient is two, but its intercept differs from negative seven, so the lines are distinct and parallel.

Answer

$y = 2 x - 2$

EXAMPLE 02

A perpendicular line

Find the line perpendicular to $y = 2 x + 1$ through (4,-1).

Worked solution

1
Find the perpendicular gradient.
$m_{⟂} = - \frac{1}{2}, 2 (- \frac{1}{2}) = - 1$
The negative reciprocal makes the product of the two gradients negative one.
2
Use the required point.
$y + 1 = - \frac{1}{2} (x - 4)$
Subtracting the given output negative one produces y plus one.
3
Simplify and verify the point.
$\begin{aligned} y & = - \frac{x}{2} + 1 \\ - \frac{4}{2} + 1 & = - 1 \end{aligned}$
The equation has the perpendicular gradient and gives the required output at input four.

Answer

$y = - x / 2 + 1$

03 / YOUR TURN

Try it yourself.

Check your understanding

What gradient is perpendicular to -3/4?

Show answer & reasoning

$4 / 3$ .

The negative reciprocal of $- \frac{3}{4}$ is $\frac{4}{3}$ . The product $(- \frac{3}{4}) (\frac{4}{3}) = - 1$ confirms the perpendicular relationship.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

Are $y = 3 x + 2$ and $6 x - 2 y = 5$ parallel, perpendicular or neither?
2 marks
Find a line parallel to $y = - 4 x + 9$ through $(1, 2)$ .
2 marks
Find a line perpendicular to $y = \frac{x}{3} - 2$ through $(2, 5)$ .
2 marks
Find the line perpendicular to $y = 7$ through $(- 2, 3)$ .
2 marks
Determine whether the lines joining $(0, 0)$ to $(2, 6)$ , and $(1, 4)$ to $(7, 2)$ , are perpendicular.
2 marks
Two paths have equations $y = 2 x + 5$ and $4 x - 2 y = - 10$ . Are they distinct parallel paths?
2 marks

Read the worked solutions

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

Are $y = 3 x + 2$ and $6 x - 2 y = 5$ parallel, perpendicular or neither?
2 marks
The second is $y = 3 x - 2.5$ : same gradient, different intercept.
Parallel and distinct.
Find a line parallel to $y = - 4 x + 9$ through $(1, 2)$ .
2 marks
$y - 2 = - 4 (x - 1)$ .
$y = - 4 x + 6$
Find a line perpendicular to $y = \frac{x}{3} - 2$ through $(2, 5)$ .
2 marks
Perpendicular gradient -3; $y - 5 = - 3 (x - 2)$ .
$y = - 3 x + 11$
Find the line perpendicular to $y = 7$ through $(- 2, 3)$ .
2 marks
The perpendicular to a horizontal line is vertical through the supplied x-coordinate.
$x = - 2$
Determine whether the lines joining $(0, 0)$ to $(2, 6)$ , and $(1, 4)$ to $(7, 2)$ , are perpendicular.
2 marks
The gradients are 3 and $\frac{- 1}{3}$ , whose product is -1.
Yes.
Two paths have equations $y = 2 x + 5$ and $4 x - 2 y = - 10$ . Are they distinct parallel paths?
2 marks
Rearranging the second gives $y = 2 x + 5$ exactly.
No; they are the same line.

Checking saved progress…

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.

Parallel lines

Perpendicular lines

See the method in action.

A parallel line

Read the original gradient after rearranging.

Use the same gradient through the required point.

Simplify and distinguish the lines.

A perpendicular line

Find the perpendicular gradient.

Use the required point.

Simplify and verify the point.

Try it yourself.

Check your understanding

Keep practising