The shape of the lesson
By the end of the lesson, learners can
- Gradients that multiply to −1: In the lab, rotate the first line and watch the perpendicular line turn with it, always at \(90^\circ\).
- Perpendicular through a point: Find the perpendicular gradient first, then substitute the given point into \(y = mx + c\) to find \(c\), exactly as for parallel lines.
- The perpendicular bisector: The perpendicular bisector of \(AB\) cuts \(AB\) in half at right angles.
- Gradient: write the given line as \(y = mx + c\) and read \(m\).
- Perpendicular: new gradient \(-\dfrac{1}{m}\); check the product is \(-1\).
- Through a point: substitute it into \(y = mx + c\) to find \(c\).
- Bisector of \(AB\): midpoint of \(AB\) plus the gradient perpendicular to \(AB\).
- Tidy: clear fractions if asked for the form \(ax + by = c\).
One picture every idea returns to
If two lines are perpendicular, \(m_1 \times m_2 = -1\), so the perpendicular gradient is \(-\dfrac{1}{m}\): flip the fraction and change the sign. The perpendicular bisector of \(AB\) passes through the midpoint of \(AB\) with the perpendicular gradient.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Flip it and change the sign.. Extended. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Gradients that multiply to −1 | In the lab, rotate the first line and watch the perpendicular line turn with it, always at \(90^\circ\). Project tab 5 of the lab (Straight lines lab, perpendicular lines) and let a student drive it while the class predicts. Model the worked example Perpendicular gradients on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Perpendicular through a point | Find the perpendicular gradient first, then substitute the given point into \(y = mx + c\) to find \(c\), exactly as for parallel lines. Model the worked example From y = mx + c on the board, one line at a time. Your turn: Find the equation of the line perpendicular to \(y = -2x + 7\) that passes through \((4, 5)\). | Pairs |
| 21 to 30 min | Teach and check: The perpendicular bisector | The perpendicular bisector of \(AB\) cuts \(AB\) in half at right angles. Model the worked example A bisector through a whole-number midpoint on the board, one line at a time. | Round Table |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 19 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in The perpendicular bisector, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Doing only half the recipe. | The perpendicular to gradient 3 is not \(-3\) (sign only) and not \(\dfrac{1}{3}\) (flip only). It is \(-\dfrac{1}{3}\). Multiply your two gradients: you must get \(-1\). |
| Using the original gradient by mistake when you substitute. | Write the NEW gradient down on its own line before finding \(c\), so you cannot slip back to the old one. |
| Putting \(A\) or \(B\) into the equation instead of the midpoint. | The bisector passes through \(M\), not through the end points, so substitute the midpoint to find \(c\). |
Perpendicular lines and bisectors: worksheet
6 questions, 19 marks. Written in the style of the papers; no past paper question is reproduced.
Write down the gradient of a line perpendicular to
(a) \(y = 5x - 1\) (b) \(y = 3 - \dfrac{3}{4}x\).
Show that the lines \(y = 4x - 3\) and \(x + 4y = 12\) are perpendicular.
Find the equation of the line perpendicular to \(x + 3y = 6\) that passes through \((2, -1)\).
The vertices of triangle \(PQR\) are \(P(-1, 1)\), \(Q(3, 3)\) and \(R(1, 7)\).
Show that angle \(PQR\) is a right angle.
\(A\) is the point \((1, 7)\) and \(B\) is the point \((7, -1)\).
(a) Find the equation of the perpendicular bisector of \(AB\).
(b) Show that the perpendicular bisector passes through the origin.
On a map with a 1 km grid, a straight road follows the line \(y = 2x + 1\). A house is at \(H(7, 0)\). A path is to be built from \(H\) to the road, as short as possible, so it must meet the road at right angles.
(a) Find the equation of the path.
(b) Find the point where the path meets the road.
(c) Find the length of the path, correct to 3 significant figures.
Answer key and marking notes
- (a) \(-\dfrac{1}{5}\)
- (b) the gradient is \(-\dfrac{3}{4}\), so the perpendicular gradient is \(\dfrac{4}{3}\).
- \(x + 4y = 12\) gives \(y = -\dfrac{1}{4}x + 3\), gradient \(-\dfrac{1}{4}\).
- \(4 \times \left(-\dfrac{1}{4}\right) = -1\), so the lines are perpendicular.
- \(y = -\dfrac{1}{3}x + 2\), gradient \(-\dfrac{1}{3}\), so the perpendicular gradient is 3.
- \(-1 = 3 \times 2 + c\), so \(c = -7\).
- \(y = 3x - 7\)
- Gradient \(PQ = \dfrac{3 - 1}{3 - (-1)} = \dfrac{2}{4} = \dfrac{1}{2}\).
- Gradient \(QR = \dfrac{7 - 3}{1 - 3} = \dfrac{4}{-2} = -2\).
- \(\dfrac{1}{2} \times (-2) = -1\), so \(PQ \perp QR\) and angle \(PQR = 90^\circ\).
- (a) Midpoint \(= (4, 3)\). Gradient \(AB = \dfrac{-1 - 7}{7 - 1} = -\dfrac{4}{3}\), so the perpendicular gradient is \(\dfrac{3}{4}\).
- \(3 = \dfrac{3}{4} \times 4 + c = 3 + c\), so \(c = 0\): \(y = \dfrac{3}{4}x\) (or \(3x - 4y = 0\)).
- (b) \(c = 0\), so when \(x = 0\), \(y = 0\): the line passes through \((0, 0)\).
- (a) Perpendicular gradient \(-\dfrac{1}{2}\). \(0 = -\dfrac{1}{2} \times 7 + c\), so \(c = 3.5\): \(y = -\dfrac{1}{2}x + 3.5\).
- (b) \(2x + 1 = -\dfrac{1}{2}x + 3.5\), so \(2.5x = 2.5\) and \(x = 1\); \(y = 2 \times 1 + 1 = 3\). The point is \((1, 3)\).
- (c) Length \(= \sqrt{(7 - 1)^2 + (0 - 3)^2} = \sqrt{36 + 9} = \sqrt{45} = 6.71\) km (3 s.f.).
6 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Find the gradient of a line perpendicular to \(y = -\dfrac{2}{5}x + 1\).
- A\(\dfrac{5}{2}\)
- B\(-\dfrac{5}{2}\)
- C\(\dfrac{2}{5}\)
- D\(-\dfrac{2}{5}\)
Find the equation of the line perpendicular to \(y = 2x + 1\) that passes through \((4, 3)\).
- A\(y = -\dfrac{1}{2}x + 5\)
- B\(y = 2x - 5\)
- C\(y = \dfrac{1}{2}x + 1\)
- D\(y = -\dfrac{1}{2}x + 3\)
Which of these lines is perpendicular to \(3x + 2y = 6\)?
- A\(2x - 3y = 5\)
- B\(3x + 2y = 1\)
- C\(2x + 3y = 5\)
- D\(3x - 2y = 4\)
Find the equation of the perpendicular bisector of \(AB\), where \(A(-1, 2)\) and \(B(5, 6)\).
- A\(y = -\dfrac{3}{2}x + 7\)
- B\(y = \dfrac{2}{3}x + \dfrac{8}{3}\)
- C\(y = -\dfrac{3}{2}x + \dfrac{1}{2}\)
- D\(y = -\dfrac{3}{2}x + \dfrac{27}{2}\)
Triangle \(PQR\) has vertices \(P(-1, 1)\), \(Q(2, 3)\) and \(R(4, 0)\). At which vertex is there a right angle?
- AAt \(Q\)
- BAt \(P\)
- CAt \(R\)
- DThere is no right angle
The lines \(2x - 5y = 3\) and \(y = kx - 1\) are perpendicular. Find \(k\).
- A\(-\dfrac{5}{2}\)
- B\(\dfrac{5}{2}\)
- C\(-\dfrac{2}{5}\)
- D\(\dfrac{2}{5}\)
Quiz key
The gradient is \(-\dfrac{2}{5}\). Flip it to \(-\dfrac{5}{2}\) and change the sign: \(\dfrac{5}{2}\). Check: \(-\dfrac{2}{5} \times \dfrac{5}{2} = -1\). The tempting \(-\dfrac{5}{2}\) flips but forgets to change the sign (the product is \(+1\)). \(\dfrac{2}{5}\) changes the sign only, and \(-\dfrac{2}{5}\) is the same gradient, which gives a parallel line.
Perpendicular gradient \(= -\dfrac{1}{2}\). Then \(3 = -\dfrac{1}{2} \times 4 + c = -2 + c\), so \(c = 5\): \(y = -\dfrac{1}{2}x + 5\). The tempting \(y = 2x - 5\) slips back to the original gradient 2 when substituting, which gives a parallel line. \(y = \dfrac{1}{2}x + 1\) flips the gradient but keeps the sign, and \(y = -\dfrac{1}{2}x + 3\) uses the point's \(y\)-coordinate as the intercept.
\(3x + 2y = 6\) gives \(y = -\dfrac{3}{2}x + 3\), gradient \(-\dfrac{3}{2}\), so a perpendicular line has gradient \(\dfrac{2}{3}\). \(2x - 3y = 5\) gives \(y = \dfrac{2}{3}x - \dfrac{5}{3}\): gradient \(\dfrac{2}{3}\), and \(-\dfrac{3}{2} \times \dfrac{2}{3} = -1\). The tempting \(2x + 3y = 5\) has gradient \(-\dfrac{2}{3}\) (flipped but same sign). \(3x + 2y = 1\) is parallel, and \(3x - 2y = 4\) has gradient \(\dfrac{3}{2}\) (sign changed only).
Midpoint \(M = \left(\dfrac{-1 + 5}{2}, \dfrac{2 + 6}{2}\right) = (2, 4)\). Gradient \(AB = \dfrac{4}{6} = \dfrac{2}{3}\), so the perpendicular gradient is \(-\dfrac{3}{2}\). Through \(M\): \(4 = -3 + c\), \(c = 7\). The tempting \(y = -\dfrac{3}{2}x + \dfrac{1}{2}\) substitutes \(A\) instead of the midpoint, giving a perpendicular through \(A\), not a bisector (\(\dfrac{27}{2}\) does the same with \(B\)). \(y = \dfrac{2}{3}x + \dfrac{8}{3}\) uses the gradient of \(AB\) itself.
Gradient \(PQ = \dfrac{2}{3}\), gradient \(QR = \dfrac{0 - 3}{4 - 2} = -\dfrac{3}{2}\), gradient \(PR = \dfrac{-1}{5}\). \(\dfrac{2}{3} \times \left(-\dfrac{3}{2}\right) = -1\), so \(PQ\) is perpendicular to \(QR\), and these two sides meet at \(Q\). The tempting "at \(P\)" picks the first vertex listed, but the sides through \(P\) have gradients \(\dfrac{2}{3}\) and \(-\dfrac{1}{5}\), whose product is not \(-1\). The angle at \(R\) is not \(90^\circ\) either.
\(-5y = -2x + 3\), so \(y = \dfrac{2}{5}x - \dfrac{3}{5}\), gradient \(\dfrac{2}{5}\). Then \(\dfrac{2}{5} \times k = -1\), so \(k = -\dfrac{5}{2}\). The tempting \(\dfrac{5}{2}\) flips the fraction but forgets the sign change. \(-\dfrac{2}{5}\) changes the sign only, and \(\dfrac{2}{5}\) is the gradient of a parallel line.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.