At a glance. One lesson of about 60 minutes. Students test pairs of gradients, write the equation of a parallel or perpendicular line through a point, and use the perpendicular condition in a general form equation and a coordinate problem.
Calculator: no calculator. Nothing here needs one. The gradients are whole numbers, halves and simple fractions.
What students will learn
Know
- Parallel lines have equal gradients.
- Perpendicular lines have gradients with product −1: m2 = −1/m1.
- A horizontal and a vertical line are perpendicular, but the product rule does not apply.
Understand
- The negative reciprocal needs the flip and the sign change; each alone fails the product test.
- A parallel line differs only in c; a perpendicular line has a different gradient and a c found by substituting.
- A product rule is a test: multiply and check for −1.
Do
- Decide whether two lines are parallel, perpendicular or neither.
- Write the equation of a parallel or perpendicular line through a given point.
- Use the perpendicular condition to find a missing value.
MYP assessment criteria
Wording paraphrased. What this lesson gives you evidence for, and what to look for:
- Criterion A. Knowing and understanding: gradient tests and equations (questions 1 to 9).
- Criterion B. Investigating patterns: the first figure. Ask students to find at least five pairs of gradients giving perpendicular lines and state the rule.
- Criterion C. Communicating: explaining why a flip alone does not work, and setting out a perpendicular equation with a check.
- Criterion D. Applying mathematics in real-life contexts: a road and a cycle lane on a map (the hook and question 6).
The one thing to do with the figure
Open on 2 and 1/2 and ask for a verdict. Many will say perpendicular, because the numbers are reciprocals. The figure says neither, with a product of 1. Then ask for the change that makes it perpendicular, and have them predict the second gradient before moving the slider.
Use the second figure to say the method aloud: negative reciprocal, substitute, check.
A possible order
| 5 min | Warm-up from the student page, in silence, then answers. |
|---|---|
| 10 min | Hook: the road, the lane and the footpath; collect predictions. |
| 15 min | First figure, beginning with 2 and 1/2. |
| 10 min | Second figure; key idea; the worked example. |
| 5 min | Question 1, the check. |
| 10 min | Your turn, Levels 1 to 3. |
| 5 min | Exit ticket, question 9. |
The answers
| Warm-up | −3; 3; 4/3. |
|---|---|
| Worked example | y = −x/3 + 6. |
| Faded example | y = −2x + 6. |
| 1. A parallel line | C. A changes the sign; B flips; D is the perpendicular. |
| 2. c for a parallel line | −5. y = 4x − 5. |
| 3. Perpendicular gradient | −0.2. −1/5. |
| 4. Perpendicular gradient of a negative fraction | 4. −1 ÷ (−0.25). |
| 5. c for a perpendicular line | 3. y = −0.25x + 3. |
| 6. A perpendicular lane on a map | 9. y = −2x + 9. |
| 7. A perpendicular condition on a general form line | 2. −2/k = −1. |
| 8. A perpendicular to find a coordinate | 7. (7 − 5) ÷ (4 − 5) = −2. |
| 9. Which pair is perpendicular? | C. A changes only the sign; B flips only; D is parallel. |
What each wrong answer tells you
| They give | What it usually means | What to do |
|---|---|---|
| A, B or D in question 1 | Used the sign change, the flip, or the perpendicular. | Ask which of the three keeps the gradient the same. |
| 3, 1 or 11 in question 2 | Used the point's y, the old c, or added. | Substitute line by line: 3 = 4 × 2 + c. |
| −5 or 0.2 in question 3; 0.25 or −4 in question 4 | Did only one of flip and sign change. | Multiply their answer by the gradient and ask if it is −1. |
| −31, 33 or −1 in question 5 | Used the parallel gradient, the sign-changed gradient, or lost a sign. | Write the new gradient first, then substitute. |
| −2, 1 or 4 in question 6 | Gave the gradient, lost a sign, or used the parallel gradient. | Ask what the question is asking for: c. |
| −2, 4.5 or 3 in question 7 | Mis-signed or inverted the second gradient. | Write each equation as y = … first. |
| 3 or 4.5 in question 8 | Lost the sign of the run, or used a parallel. | Draw the points and mark the run from B to C. |
| A, B or D in question 9 | Checked sign only, flip only, or used a parallel. | Multiply the two gradients in each option. |
Questions worth asking
- Which of two lines is steeper, gradient 3 or −1/3? Gradient 3. They are perpendicular: one is steep, the other is shallow and falls.
- If two lines are perpendicular, can both have positive gradients? No: the product is negative, so exactly one is negative.
- Can two parallel lines have different c? They must, unless they are the same line.
- How can you check a perpendicular line? Multiply the gradients to get −1, and check that the point fits.
Differentiation
A way in for anyone stuck
- Use a table: given gradient, flipped, sign changed, product. Fill in all four every time.
- Start with whole-number gradients and parallel lines before perpendicular.
- Provide a template for equations: m = …, put the point in, c = …, equation.
Stretch
- Question 7, then: show that the lines ax + by = d and bx − ay = e are always perpendicular.
- Find the equation of the perpendicular bisector of the segment from (0, 0) to (6, 4). (y = −1.5x + 6.5.)
Students learning in English as an additional language
“Parallel” means going the same way and never meeting. “Perpendicular” means meeting at a right angle, like the corner of a page. “Reciprocal” means a number flipped upside down.
| Word | Plain meaning | Example |
|---|---|---|
| parallel | never meeting, with equal gradients | y = 2x + 1 and y = 2x − 4 |
| perpendicular | meeting at a right angle | gradients 2 and −1/2 |
| reciprocal | a number flipped: 1 divided by it | the reciprocal of 4 is 1/4 |
| negative reciprocal | the reciprocal with the opposite sign | the negative reciprocal of 2 is −1/2 |
| right angle | an angle of 90 degrees | the corner of a square |
| product | the answer to a multiplication | the product of 2 and −1/2 is −1 |
Sentence starters for explaining aloud or in writing:
- The gradients are … and …, and their product is …
- The lines are parallel because …
- The new gradient is the negative reciprocal, which is …
Two things not to say
“Flip it and change the sign” with no reason. Always give the product −1 as the reason.
“Perpendicular lines have opposite gradients.” Opposite is the sign change alone.
Questions to set
Extra practice in three tiers, including one question from earlier work for spacing. Every answer has been checked.
1Fluency
The idea on its own, with friendly numbers. Set these first.
- TestAre y = 3x − 1 and y = −x/3 + 2 parallel, perpendicular or neither?
Perpendicular: the product is −1. - ParallelWrite the line parallel to y = x/2 + 1 through (4, 5).
y = x/2 + 3.
2In context
The same idea inside a situation, where the first job is deciding what to calculate.
- ContextA fence runs along y = 4x on a plan. A gate path is perpendicular to it through (8, 1). Write its equation.
y = −x/4 + 3. - Earlier work (Lesson 5)Find the gradient of 6x + 3y = 9.
−2.
3Challenge
Reasoning, generalizing and working backwards.
- ReasoningIs the triangle with vertices A(0, 0), B(4, 2) and C(1, 6) right-angled? Where?
AB has gradient 1/2 and AC has gradient 6, BC has gradient −4/3; no pair multiplies to −1, so no. - ReasoningThe lines y = mx + 1 and y = (m + 2)x − 3 are parallel. What is wrong?
Their gradients differ by 2, so they cannot be parallel for any m.
Practicalities
Works on a phone or a shared screen. Nothing a student types is saved or sent anywhere.
Both figures work with the keyboard (tab to a slider, then the arrow keys) and by touch. Under each one, “Show me the steps” gives the same states in words.
No device? Use squared paper and a set square to test the right angle.