The shape of the lesson
By the end of the lesson, learners can
- Rise over run: Drag the two points in the lab and watch the gradient triangle.
- Two points, one formula: (Extended) When there is no grid, use the coordinates.
- What the gradient tells you: A gradient is a rate: how much \(y\) changes for each 1 unit of \(x\).
- Grid: draw a right-angled triangle under the line between two exact grid points.
- Coordinates (Extended): \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\), same order top and bottom.
- Sign: uphill positive, downhill negative.
- Special lines: horizontal 0, vertical undefined.
- Meaning: the gradient is the change in \(y\) for each 1 unit of \(x\), with units \(y\)-units per \(x\)-unit.
One picture every idea returns to
Gradient \(= \dfrac{\text{rise}}{\text{run}} = \dfrac{\text{change in } y}{\text{change in } x}\). Uphill from left to right is positive, downhill is negative, a horizontal line has gradient 0, and a vertical line has no gradient (it is undefined).
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, How steep is steep?. The gradient of a line is one number that measures its steepness and its direction: how far it goes up for every one step across. 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: Rise over run | Drag the two points in the lab and watch the gradient triangle. Project tab 1 of the lab (Straight lines lab, gradient) and let a student drive it while the class predicts. Model the worked example Uphill line from a grid on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Two points, one formula | (Extended) When there is no grid, use the coordinates. Model the worked example Negative coordinates (Extended) on the board, one line at a time. Your turn: (Extended) Find the gradient of the line through \((-3, -4)\) and \((1, 8)\). | Pairs |
| 21 to 30 min | Teach and check: What the gradient tells you | A gradient is a rate: how much \(y\) changes for each 1 unit of \(x\). Model the worked example Gradient as a rate on the board, one line at a time. | Numbered Heads Together |
| 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, 16 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 What the gradient tells you, 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.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
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 |
|---|---|
| Counting squares when the axes have different scales. | If one square across is 2 units and one square up is 10 units, read the actual values off the axes, not the number of squares. |
| Taking the \(y\) values in one order and the \(x\) values in the other order. | If you start with \(y_2\) on top, you must start with \(x_2\) underneath, or the sign flips. And it is \(y\) over \(x\), never \(x\) over \(y\). |
| A horizontal line has gradient 0 (no rise at all), but a VERTICAL line does not have gradient 0: its run is 0, and you cannot divide by 0, so its gradient is undefined. | Return to What the gradient tells you: ask a pair to state the correct version and show it on the lab before moving on. |
Gradient of a straight line: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
State whether the gradient of each line is positive, negative, zero or undefined.
(a) \(y = 5\) (b) \(x = -2\) (c) \(y = 3x\) (d) \(y = 6 - x\)
A straight line on a grid passes through the points \((-1, -4)\) and \((3, 8)\). Find its gradient.
(Extended) Calculate the gradient of the line joining \((-4, 6)\) and \((6, 1)\).
(Extended) Show that the points \(A(-1, 1)\), \(B(1, 5)\) and \(C(4, 11)\) lie on a straight line.
(Extended) The line through \((3, -2)\) and \((t, 10)\) has gradient 4. Find \(t\).
A travel graph for a car journey has two straight sections. The first joins \((0, 0)\) to \((2.5, 200)\) and the second joins \((2.5, 200)\) to \((4, 260)\), where \(x\) is the time in hours and \(y\) is the distance in km.
(a) Find the gradient of each section.
(b) What do the gradients represent? In which section was the car faster?
Answer key and marking notes
- (a) zero (horizontal)
- (b) undefined (vertical)
- (c) positive
- (d) negative (\(y\) falls as \(x\) increases)
- Run: from \(-1\) to \(3\) is 4. Rise: from \(-4\) to \(8\) is 12.
- Gradient \(= 12 \div 4 = 3\).
- \(m = \dfrac{1 - 6}{6 - (-4)} = \dfrac{-5}{10}\)
- \(m = -\dfrac{1}{2}\)
- Gradient \(AB = \dfrac{5 - 1}{1 - (-1)} = \dfrac{4}{2} = 2\).
- Gradient \(BC = \dfrac{11 - 5}{4 - 1} = \dfrac{6}{3} = 2\).
- The gradients are equal and the two segments share the point \(B\), so \(A\), \(B\) and \(C\) lie on one straight line.
- \(\dfrac{10 - (-2)}{t - 3} = 4\), so \(\dfrac{12}{t - 3} = 4\).
- \(12 = 4(t - 3) = 4t - 12\), so \(4t = 24\).
- \(t = 6\).
- (a) First: \(\dfrac{200 - 0}{2.5 - 0} = 80\). Second: \(\dfrac{260 - 200}{4 - 2.5} = \dfrac{60}{1.5} = 40\).
- (b) Each gradient is a speed in km/h (km per hour).
- The first section is steeper, so the car was faster there: 80 km/h compared with 40 km/h.
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 the straight line in the diagram.
- A\(-2\)
- B\(2\)
- C\(-\dfrac{1}{2}\)
- D\(4\)
Which of these lines has a gradient of 0?
- A\(y = 4\)
- B\(x = 4\)
- C\(y = 4x\)
- D\(x + y = 4\)
A taxi fare is shown by a straight-line graph from \((0, 3.5)\) to \((8, 15.5)\), where \(x\) is the distance in km and \(y\) is the fare in dollars. What does the gradient tell you?
- AEach extra km costs $1.50
- BEach extra km costs $1.94
- CEach extra km costs $0.67
- DEach extra km costs $12
Find the gradient of the line joining \((-4, 3)\) and \((2, -6)\).
- A\(-\dfrac{3}{2}\)
- B\(\dfrac{3}{2}\)
- C\(-\dfrac{2}{3}\)
- D\(-\dfrac{9}{2}\)
The line through \((k, 2)\) and \((4, 10)\) has gradient 4. Find \(k\).
- A\(2\)
- B\(6\)
- C\(-2\)
- D\(8\)
The points \(A(1, 2)\), \(B(3, 8)\) and \(C(6, k)\) lie on the same straight line. Find \(k\).
- A\(17\)
- B\(18\)
- C\(11\)
- D\(26\)
Quiz key
The marked points are \((0, 4)\) and \((3, -2)\). The run is 3 and the line goes DOWN 6, so the rise is \(-6\) and the gradient is \(\dfrac{-6}{3} = -2\). The tempting 2 forgets that a downhill line has a negative gradient. \(-\dfrac{1}{2}\) divides run by rise (upside down), and 4 is the \(y\)-intercept, not the gradient.
\(y = 4\) is horizontal: \(y\) stays 4 whatever \(x\) is, so the rise is 0 and the gradient is 0. The tempting \(x = 4\) is VERTICAL: its run is 0, and you cannot divide by 0, so its gradient is undefined, not 0. \(y = 4x\) has gradient 4, and \(x + y = 4\) is \(y = 4 - x\), with gradient \(-1\).
Rise \(= 15.5 - 3.5 = 12\) dollars; run \(= 8 - 0 = 8\) km. Gradient \(= 12 \div 8 = 1.5\) dollars per km. The tempting $1.94 is \(15.5 \div 8\): it ignores the $3.50 starting charge, so it is not the change in fare. $0.67 is \(8 \div 12\) (run over rise), and $12 is the rise alone, not divided by the run.
Change in \(y = -6 - 3 = -9\); change in \(x = 2 - (-4) = 6\). Gradient \(= \dfrac{-9}{6} = -\dfrac{3}{2}\). The tempting \(\dfrac{3}{2}\) takes the \(y\) values in one order and the \(x\) values in the other, which flips the sign. \(-\dfrac{2}{3}\) is \(x\) over \(y\), and \(-\dfrac{9}{2}\) works out \(2 - (-4)\) as 2.
\(\dfrac{8}{4 - k} = 4\), so \(4(4 - k) = 8\), \(4 - k = 2\) and \(k = 2\). Check: \(\dfrac{10 - 2}{4 - 2} = 4\). The tempting 6 uses \(k - 4\) on the bottom with \(10 - 2\) on top (mixed order), giving gradient \(-4\) instead. \(-2\) uses \(4 + k\), and 8 forgets to multiply the \(k\) by 4: \(16 - k = 8\).
Gradient of \(AB = \dfrac{8 - 2}{3 - 1} = 3\). From \(B\) to \(C\) the run is \(6 - 3 = 3\), so the rise is \(3 \times 3 = 9\) and \(k = 8 + 9 = 17\). The tempting 18 is \(3 \times 6\): it assumes the line passes through the origin (\(y = 3x\)), but the line is \(y = 3x - 1\). 11 adds only one gradient step to 8, and 26 multiplies the gradient by the \(x\)-coordinate 6 instead of by the run 3.
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.