At a glance. One lesson of about 60 minutes. It assumes gradients and the equation of a line. Students rotate points about the origin and another centre, find perpendicular gradients and lines, and find a perpendicular bisector.
Calculator: no calculator. The numbers are small and exact and the aim is structure, so no calculator is used.
What students will learn
Know
- (a, b) goes to (−b, a), (b, −a) or (−a, −b) under 90° anticlockwise, 90° clockwise or 180°.
- Perpendicular gradients multiply to −1.
- A line through a point with a gradient: y − y₁ = m(x − x₁).
Understand
- A quarter turn keeps the radius and turns it through a right angle.
- A rotation about C is a rotation about the origin carried out in the frame of C.
- A perpendicular bisector needs a midpoint and the negative reciprocal.
Do
- Rotate a point about the origin or another point.
- Find a perpendicular gradient and write a perpendicular line as ax + by + c = 0.
- Find a perpendicular bisector.
MYP assessment criteria
Wording paraphrased. What this lesson gives you evidence for, and what to look for:
- Criterion A. Knowing and understanding: rotation and perpendicular lines (questions 2 to 8).
- Criterion B. Investigating patterns: the what-if, what a half turn does to a line.
- Criterion C. Communicating: equations written in the form ax + by + c = 0 with every step shown.
- Criterion D. Applying mathematics in real-life contexts: the map on the table, and a bisector for the position of a meeting point.
The one thing to do with the figure
Ask for the image and the gradient before moving the slider. Set P to (4, 3) and ask for the image of a quarter turn, then for the gradient of the new line, before the figure's text is read.
Then set P on an axis and ask what happens to the product of the gradients, and why the rule fails.
A possible order
| 5 min | Warm-up from the student page, in silence, then answers. |
|---|---|
| 5 min | Hook: the map. Collect predictions for the landmark's new position. |
| 12 min | The figure with predictions; the key idea and the gradient argument; settle the hook. |
| 10 min | Warning card and worked examples; pairs finish the faded example. |
| 5 min | Question 1, the check. If many choose B, return to which way the turn goes. |
| 18 min | Your turn, Levels 1 to 3, at each student's own pace. |
| 5 min | Something to think about, then the exit ticket, question 9. |
The answers
| Warm-up | 3; (2, 5); y = 2x − 3. |
|---|---|
| Hook | The landmark at (5, 2) goes to (−2, 5), and the new line has gradient −5/2, perpendicular to the first. |
| Worked examples | (−7, −4); 2x − 3y − 15 = 0; (−4, 6). |
| Faded example | x + 3y − 9 = 0. |
| 1. Quarter turn anticlockwise | A. B is clockwise, C a half turn, D a swap with no sign change. |
| 2. Clockwise quarter turn | −10. (2, −10). |
| 3. Perpendicular gradient | 1.25. 5/4. |
| 4. Perpendicular gradient from general form | 2.5. 5/2. |
| 5. Intercept of a perpendicular line | 8. y = −(3/2)x + 8. |
| 6. A constant in a perpendicular line | −6. 4 + 2 + c = 0. |
| 7. Rotation about another centre | −1. Image (−1, 8). |
| 8. Perpendicular bisector | −4.5. y = (3/2)x − 4.5. |
| 9. Choosing a perpendicular line | D. Gradients 3/4 and −4/3 multiply to −1. |
What each wrong answer tells you
| They give | What it usually means | What to do |
|---|---|---|
| B, C or D in question 1 | Turned the wrong way, half a turn, or swapped without changing a sign. | Check with a sketch of the quadrants. |
| 10 or −2 in question 2 | Turned anticlockwise, or half a turn. | Write the rule before using it. |
| −1.25, 0.8 or −0.8 in question 3 | Kept the sign, dropped the sign, or gave the same gradient. | Say “flip and change the sign”. |
| −2.5 or 0.4 in question 4 | Read the gradient with the wrong sign, or did not take the reciprocal. | Rearrange to y = mx + c first. |
| 2 or 5 in question 5 | Subtracted instead of adding, or gave the point's y. | Put x = 0 in the equation. |
| 6 or −9 in question 6 | Sign slip, or swapped the coordinates. | Substitute x then y with the point in a bracket. |
| −4 or −3 in question 7 | Turned about the origin, or forgot to add the centre. | Three steps: subtract, turn, add. |
| 4.5, 3 or 6.3 in question 8 | Sign slip, gave the midpoint's y, or used the gradient of AB. | Draw A, B and the midpoint on a sketch. |
| A, B or C in question 9 | Negated, reciprocated only, or gave a parallel line. | Find both gradients and multiply. |
Questions worth asking
- Why does a quarter turn give gradients with product −1? (a, b) goes to (−b, a): b ÷ a times a ÷ (−b) is −1.
- What happens to the gradient of a horizontal line under a quarter turn? It becomes vertical, with no gradient.
- How can a quarter turn be undone? By a quarter turn the other way, or three more in the same direction.
- Is the perpendicular bisector unique? Yes: it goes through the midpoint and has the perpendicular gradient.
Differentiation
A way in for anyone stuck
- Use tracing paper and a pin at the origin to turn a point.
- Give a table: a, b, anticlockwise, clockwise, half turn.
- Keep the first equations in the form y = mx + c before writing ax + by + c = 0.
Stretch
- Show that rotating the line y = mx about the origin through 90° gives the line y = −x ÷ m.
- Find the centre of the rotation of 90° anticlockwise that sends (4, 1) to (2, 5).
Students learning in English as an additional language
“Perpendicular” means meeting at a right angle. “Gradient” is how steep a line is. “Clockwise” is the way a clock's hands move and “anticlockwise” is the other way. A “bisector” cuts something into two equal parts. Show the turns with your hand.
| Word | Plain meaning | Example |
|---|---|---|
| quarter turn | a rotation of 90° | from facing north to facing east is a clockwise quarter turn |
| anticlockwise | the direction opposite to the hands of a clock | the direction positive angles are measured |
| perpendicular | meeting at a right angle | the two sides of a set square at its corner |
| negative reciprocal | flip the fraction and change its sign | −7 ÷ 2 becomes 2 ÷ 7 |
| perpendicular bisector | the line through the midpoint of a segment at a right angle to it | the fold line when two points are folded onto each other |
Sentence starters for explaining aloud or in writing:
- The image is (…, …) because the rule for this turn is …
- The gradients multiply to −1, so the lines are …
- I subtracted the centre first because …
Two things not to say
“Swap the coordinates and change a sign.” Which sign, and for which direction?
“The gradient is just flipped.” The sign changes as well.
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.
- TurnRotate (5, 2) through 180° about the origin.
(−5, −2). - GradientFind the gradient of a line perpendicular to y = 4x + 1.
−1/4.
2In context
The same idea inside a situation, where the first job is deciding what to calculate.
- ContextA robot at (3, 4) turns 90° anticlockwise about a post at the origin. Where does it end up?
(−4, 3). - Earlier work (Year 9)A line has gradient 2 and passes through (1, 5). Where does it cross the y-axis?
(0, 3).
3Challenge
Reasoning, generalizing and working backwards.
- ReasoningFind the equation of the perpendicular bisector of (−1, 2) and (5, 6) in the form ax + by + c = 0.
3x + 2y − 14 = 0. - ProofShow that if a point is rotated 90° about the origin, its distance from the origin is unchanged.
(−b)² + a² = a² + b².
Practicalities
Works on a phone or a shared screen. Nothing a student types is saved or sent anywhere.
The figure works with the keyboard (tab to a slider, then the arrow keys) and by touch. “Show me the steps” gives the same states in words.
No calculator is needed. Every answer is reached by hand.