A quarter turn
y = 2x: 1 right, 2 up
Turned a quarter turn: 2 right, 1 down
Gradient: −1 ÷ 2 = −½
2 × (−½) = −1
Parallel and perpendicular lines Why do the gradients of perpendicular lines multiply to −1?
Higher · AQA · 45 minutes
Students guess the gradient of a line at right angles, then turn a step a quarter turn to see why it is −1/m. Slides to present, a vote that shows who's stuck, practice and homework.
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Move m. The two lines always cross at a right angle. Multiply their gradients each time: what do you get?
y = 2x + 1 and y = −(1/2)x + 1
The idea in one minute
Parallel lines have the same gradient. Perpendicular lines cross at a right angle, and their gradients multiply to −1: a line with gradient m has perpendicular partners with gradient −1/m. To find one, flip the gradient and change its sign. Turning a line's step of 1 right and m up a quarter turn gives a step of m right and 1 down, which is why.
The lesson goes picture first. Students guess a perpendicular gradient before anyone explains, then turn a step on the graph and move a slider to see the product stay −1. Only then do they find whole lines through a point, parallel or perpendicular, and rearrange equations before reading a gradient. It is Higher; the stretch finds where such a line meets the x-axis.
By the end, students can
- say why parallel lines have the same gradient
- find a perpendicular gradient by flipping and changing the sign, also for fractions
- test two lines for perpendicular by multiplying their gradients
- find the equation of a line through a point, parallel or perpendicular to another
Key words
- parallel
- perpendicular
- gradient
- negative reciprocal
- y = mx + c
Before this lesson
Students should read m and c from y = mx + c, find a gradient from two points and rearrange an equation into y = mx + c (the lesson Straight-line graphs).
Taught in Straight-line graphs.
Next: The equation of a circle.
The lesson, step by step
30 minutes in total. Each step says what changes for this length.
3 min 1 · Hook · students think
Find the right-angle partner
Students guess the gradient of a perpendicular line before anyone explains.
For this length: Show the pair and ask for one guess; skip gathering guesses
Say“y = 2x + 1 and y = −½x + 1 cross at a right angle. Which gradient crosses y = 3x + 1 at a right angle? Write your guess and why.”
Ask- What do 2 and −½ have in common?
- Would −3 work? How could we test it?
Watch forGuessing −3 (only the sign changed): keep it. The next step tests it.
For the teacher
Give a minute of silent writing, then gather guesses. Many will say −3: one line goes up, so the other must go down by the same.
Don't explain yet. The question is the point: what links the two gradients of lines at right angles?
Gather guesses with Ask the class. A split between −3 and −⅓ makes the next step matter.
Shorter: Show the pair and ask for one guess; skip gathering guesses (−2 min).
8 min 2 · Explain · you explain
Parallel, then a quarter turn
Parallel lines share a gradient; turn a step a quarter turn to find the perpendicular one.
For this length: Skip the parallel slide's votes; one move of m
Say“A gradient is a step: one right, m up. Turn that step a quarter turn and see what gradient comes out.”
Ask- y = 2x: one step right, how far up? Turned a quarter turn, how far right and how far down?
- Before I move m to 3: what will the partner's gradient be?
Watch forChanging only the sign: test −3 against y = 3x on the graph; it does not make a right angle.
Through a point
Perpendicular to y = 2x − 3, through (4, 1)
Gradient: −½
1 = −½ × 4 + c, so c = 3
y = −½x + 3
For the teacher
First slide: y = 2x + c and y = 2x − 2. Move c: the lines stay the same steepness and never meet. Parallel means the same gradient; c only moves a line up or down.
Second slide: the step of y = 2x is 1 right, 2 up. Turned a quarter turn clockwise about (0, 0), the step's triangle has its 1 across pointing down and its 2 up pointing right: 2 right, 1 down, ending at (2, −1). Both triangles are drawn; trace the turn with a finger. So the gradient is −1 ÷ 2 = −½: flipped, and its sign changed. 2 × (−½) = −1. This answers the hook: −3 only changes the sign.
Third slide: move m. The partner −1/m always multiplies with m to give −1. At m = 0 the line is flat and its partner is upright, x = 0, which has no gradient: the rule is for lines that are neither.
Fourth slide: a whole line through a point. The gradient first, then put the point into y = mx + c to find c.
Vote before each move (Ask the class): c to −2 on the parallel slide; m to 3 on the slider slide.
Shorter: Skip the parallel slide's votes; one move of m (−4 min).
5 min 3 · Your turn · students write
Worked, then yours
Each worked example beside a twin that students solve themselves.
For this length: Do the first and third pairs only
Say“Watch me do one. Then you do its twin, which is almost the same.”
Ask- Do your two gradients multiply to −1?
Watch forFlipping a negative gradient but keeping its sign: −3 goes to ⅓, not −⅓.
A perpendicular gradient
y = 4x − 1
Flip and change the sign: −¼
4 × (−¼) = −1
Then theirs
y = −3x + 2
A fraction gradient
y = ⅔x + 1
Flip: 3/2. Change the sign: −3/2
⅔ × (−3/2) = −1
Then theirs
y = −¼x + 5
A whole line through a point
Perpendicular to y = 3x + 2, through (3, 4)
Gradient −⅓; 4 = −⅓ × 3 + c, so c = 5
y = −⅓x + 5
Then theirs
Perpendicular to y = 2x + 7, through (2, 6)
For the teacher
Work the first example aloud, writing each line. Then students solve its twin while you watch who changes the sign of a negative gradient.
Do the same for each pair. The second pair catches students who flip a fraction but keep its sign; the third, those who use the point's y as c.
Open the answers slide once most have finished their twins.
Shorter: Do the first and third pairs only (−3 min).
9 min 4 · Who's stuck? · students vote
Vote, then practise
One question shows who's stuck; then practice in levels, and a stretch.
For this length: After the vote, questions 1–4 only
Say“Choose an answer on your own. Nobody talks until everyone has chosen.”
Ask- How can you check a perpendicular gradient without a graph?
Watch forChanging the sign or flipping, but not both.
Show your working on paper. Write fractions with /, such as -1/3, and negative numbers with a minus sign. The stretch is optional.
The question that shows who's stuck
- A line has gradient −⅔. What is the gradient of a line perpendicular to it?
In your Classroom, as they answer (an example) - A10
- B8
- C5
- D2
Each wrong answer is a different mistake. While they answer, your Classroom shows who chose which, so you know in seconds whom to help with what.
- A line parallel to y = 5x − 2 has gradient ___
- A line perpendicular to y = 5x − 2 has gradient ___
- 2y = 6x + 1. A line perpendicular to it has gradient ___
- y = 4 − ½x. What is the gradient of a line perpendicular to it?
- Which pair of lines is perpendicular?
- Line A goes through (0, 1) and (2, 5). Line B goes through (0, 3) and (4, 1). Are they perpendicular?
- The line perpendicular to y = 2x + 1 through (4, 3) is y = ___x + ___
- The line parallel to y = 3x − 2 through (1, 7) is y = 3x + ___
- A line goes through (1, 2) and (5, 4). A line perpendicular to it has gradient ___
- Stretch: the line through (1, 6) perpendicular to y = ⅓x + 1 meets the x-axis at x = ___
Support · For question 3: divide every term by 2 first, so the equation starts y = … . Then flip the gradient and change its sign. Check: the two gradients multiply to −1.
For the teacher
Everyone chooses before anyone explains. If they changed only the sign, or only flipped, ask them to multiply their answer by −⅔. If they chose −⅔, ask whether that line is parallel or perpendicular.
Questions 1–2 build fluency; 3–4 change the look; 5–6 ask for reasons; 7–9 find whole lines, as exam questions do. The stretch finds where such a line meets the x-axis.
Students show their working on paper or the board; the quiz keeps their final answers.
If the class is split, multiply each option by −⅔ together before they practise: only one gives −1.
Shorter: After the vote, questions 1–4 only (−6 min).
Longer: The stretch, then: why is there no partner gradient for a flat line? (+10 min).
5 min 5 · Exit · students show
Two partners for one line
One parallel and one perpendicular line through the same point show who has both rules.
Say“Write a line parallel to L and a line perpendicular to L, both through (0, 2). Say how you know each.”
Watch forA perpendicular line written with gradient −4: only the sign changed.
Two equations and one sentence are enough.
- L is the line y = 4x − 3. Write the equation of the line parallel to L through (0, 2), and of the line perpendicular to L through (0, 2). Say how you know each one.
For the teacher
L is y = 4x − 3. Write the line parallel to L through (0, 2), and the line perpendicular to L through (0, 2).
Then say in a sentence how you know each one.
The answers show who still changes only the sign: start the next lesson with the quarter turn again. Set the homework.
Longer: Sketch all three lines and mark the right angle (+5 min).
One to one: let the student guess the partner gradient first, then test it on the graph; ask them to multiply the gradients aloud before every answer.
Homework: gradients and lines
Show your working. Rearrange into y = mx + c before reading a gradient. Check every perpendicular pair: the gradients multiply to −1.
- Write the gradient of a line parallel to y = 6x − 1, and of a line perpendicular to it.
- Write the gradient of a line perpendicular to y = −⅗x + 2.
- Write 3x + 2y = 6 in the form y = mx + c. Then write the gradient of a line perpendicular to it.
- Find the equation of the line perpendicular to y = 2x − 5 that goes through (6, 1).
- Show that y = 3x + 1 and x + 3y = 12 are perpendicular.
- A is (0, 1) and B is (2, 5). Find the equation of the line through B that is perpendicular to AB.
Print the worksheet with working space and homework. Teacher answers come with your Library copy.
Teaching parallel and perpendicular lines: common questions
Why do the gradients of perpendicular lines multiply to −1?
A gradient m is a step of 1 right and m up. Turn that step a quarter turn and it becomes m right and 1 down, a gradient of −1/m. So m × (−1/m) = −1. Drawing both steps on squared paper, or on equal axes, shows it better than the rule alone.
How do you find the perpendicular gradient of a fraction like −4/5?
Flip it and change its sign: −4/5 flips to −5/4, and the sign changes to give 5/4. Check by multiplying: −4/5 × 5/4 = −1. Students who do only one of the two get +1 or −16/25, which shows them the slip.
What about horizontal and vertical lines?
A horizontal line has gradient 0 and the line at right angles to it is vertical, such as x = 2, which has no gradient. So the rule that gradients multiply to −1 is for lines that are neither horizontal nor vertical. It is worth saying when the slider reaches m = 0.
Are parallel and perpendicular lines on the Foundation or Higher tier?
Parallel lines, as lines with the same gradient, are on both tiers. Perpendicular gradients and finding a perpendicular line through a point are Higher. The lesson builds on Straight-line graphs, and leads on to the equation of a tangent to a circle.
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Curriculum reference: AQA GCSE Mathematics 8300: A9 (parallel and perpendicular lines from y = mx + c, Higher; the equation of a line through one point with a given gradient).
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