At a glance. One lesson of about 60 minutes. Students move from a table with a step of 1 to any two points. They find the gradient as rise over run with negative coordinates, meet zero and undefined gradients, and use the gradient as a rate of change in a context.
Calculator: no calculator. Nothing here needs one. Every run divides into the rise as a whole number or a short decimal, so the work stays on the structure.
What students will learn
Know
- Gradient = rise ÷ run = (y2 − y1) ÷ (x2 − x1).
- The sign of the gradient says whether the line rises or falls; a horizontal line has gradient 0; a vertical line has none.
- A gradient in context is a rate of change with units, for example metres per year.
Understand
- Any two points on a straight line give the same gradient, because the rate of change is constant.
- A bigger rise does not mean a steeper line: the run matters as much.
- Swapping A and B changes the sign of both the rise and the run, so the gradient is unchanged.
Do
- Find the gradient from two points, including negative coordinates.
- Interpret a gradient as a rate with units.
- Use a gradient to find a missing coordinate or a further point on a line.
MYP assessment criteria
Wording paraphrased. What this lesson gives you evidence for, and what to look for:
- Criterion A. Knowing and understanding: calculating a gradient and handling signs (questions 1 to 5 and 9).
- Criterion B. Investigating patterns: the second figure. Ask students to choose five different pairs, record rise and run, and state what they notice about the ratio.
- Criterion C. Communicating: explaining why a vertical line has no gradient and why the order of subtraction must match (the key card and warning).
- Criterion D. Applying mathematics in real-life contexts: road steepness, a glacier, a mountain trail, a bicycle and a solar lamp (the hook and questions 3, 6 and the examples).
The one thing to do with the figure
Ask for the gradient before each new setting. Set A and B, ask for a prediction of the sign, then the rise and run, then show the readout. Use the fifth state in the steps list for a vertical line and ask what happens when you divide by the run.
In the second figure, give pairs of students a pair of points each and have them say, before moving, what the gradient must be. Ask why a larger triangle does not change it.
A possible order
| 5 min | Warm-up from the student page, in silence, then answers. |
|---|---|
| 10 min | Hook: the two roads. Collect predictions; the answer comes with the key idea. |
| 10 min | First figure: rise, run and gradient across the six states, with predictions first. |
| 5 min | Second figure: any two points on one line. |
| 10 min | Key idea; the mountain trail worked example; students finish the solar lamp. |
| 5 min | Question 1, the check. If many choose C or D, return to rise and run on the first figure. |
| 10 min | Your turn, Levels 1 to 3. |
| 5 min | Exit ticket, question 9. |
The answers
| Warm-up | 4; −10; −4. |
|---|---|
| Worked example | 100 m per km. |
| Faded example | −20 per cent per hour. |
| 1. Gradient from two points | B, −2. A drops the sign; C divides run by rise; D stops at the rise. |
| 2. Gradient from two points | 0.5. Rise 2 over run 4. |
| 3. Negative gradient from a rate | −5. (150 − 180) ÷ 6. |
| 4. Gradient with negative coordinates | −1.5. −9 ÷ 6. |
| 5. Gradient of a horizontal line | 0. A horizontal line has rise 0. |
| 6. Gradient as a rate | 0.35. 21 ÷ 60. |
| 7. A missing coordinate from a gradient | 9. Check: (9 − 1) ÷ (6 − 2) = 2. |
| 8. Extending a line by gradient | 14. Gradient 3; the run 2 gives a rise of 6. |
| 9. A point on a line of given gradient | B. The gradient from the origin is 1.5; only (4, 6) matches it. A is 4/3, C is 4/3, D is 1. |
What each wrong answer tells you
| They give | What it usually means | What to do |
|---|---|---|
| A in question 1; 1.5 in question 4 | Lost the sign. | Ask whether the line rises or falls, and what that says about the sign of the gradient. |
| C in question 1; 2 in question 2 | Divided run by rise. | Ask for the rise and the run in words and write “rise over run” with the rise on top. |
| D in question 1; −9 in question 4; 21 in question 6 | Stopped at the rise. | Ask: change in y for how many steps of x? The gradient is the change for one step. |
| 4.5 in question 4 | Treated the −4 as 4 in the run. | Write the subtraction with brackets: 2 − (−4). Mark it on a number line. |
| 5, 30 or −30 in question 3 | Dropped the sign, or did not divide by the 6 years. | Ask if the glacier is thicker or thinner, and what “per year” asks for. |
| 8 or 4 in question 5 | Gave the run or the y value in a horizontal line. | Ask what the change in y is between two points at the same height. |
| 8, 13 or 3 in question 7 | Gave the rise, used x as the run, or added the gradient once. | Draw the triangle with the run marked from 2 to 6. |
| 11 or 10 in question 8 | Added the gradient once, or added the run. | Ask how far across B to C is, and how much each step across raises the line. |
Questions worth asking
- Does swapping A and B change the gradient? No. Both the rise and the run change sign. Listen for “a negative divided by a negative”.
- Why does a vertical line have no gradient? The run is 0 and you cannot divide by 0. Listen for “it never goes across”.
- Which is steeper, a gradient of −3 or of 2? −3: steepness is the size of the gradient; the sign gives the direction.
- What does a gradient of 0.35 km per minute tell you? The distance rises 0.35 km every minute: a speed.
Differentiation
A way in for anyone stuck
- Give a grid and ask students to draw the triangle for every question and write the rise and run on it.
- Start with pairs where the run is 1 or 2 and the points are in the first quadrant, then add negatives.
- Provide the fraction template: change in y over change in x, with boxes for each subtraction.
Stretch
- Question 8, then: is D(7, 21) on the line through A(1, 2) and B(3, 8)? Explain with gradients. (No: from B the gradient is 13/4, not 3.)
- Find the gradient of the line through (a, 2a) and (3a, 8a) for any a not equal to 0. (3.)
Students learning in English as an additional language
“Rise” and “run” are everyday words used as technical words: say them while moving a finger up and across. “Gradient” and “slope” are used for the same thing.
| Word | Plain meaning | Example |
|---|---|---|
| gradient | how steep a line is: the rise divided by the run | a gradient of 2 rises 2 for every 1 across |
| rise | the change in y between two points | from y = 2 to y = 6 is a rise of 4 |
| run | the change in x between two points | from x = 1 to x = 5 is a run of 4 |
| slope | another word for gradient | the slope of a road |
| horizontal | flat, going straight across | a line with gradient 0 |
| vertical | straight up and down | a line with no gradient |
| undefined | has no value, because the working cannot be done | a rise over a run of 0 |
Sentence starters for explaining aloud or in writing:
- The rise is … and the run is …, so the gradient is …
- The line falls from left to right, so the gradient is …
- The gradient is undefined because …
Two things not to say
“Rise over run, flip it if it's negative.” The sign comes from the subtraction, not from a flip.
“Count the squares.” That works on a drawn grid, but it does not work for coordinates you are given.
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.
- GradientFind the gradient of the line through (0, 1) and (4, 9).
2. - SignFind the gradient of the line through (1, 7) and (4, 1).
−2.
2In context
The same idea inside a situation, where the first job is deciding what to calculate.
- ContextA tank has 80 litres at 2 hours and 20 litres at 8 hours. Find the rate of change.
−10 litres per hour. - Earlier work (Lesson 1)A linear table has y = 5 at x = 0 and the change is −3 per step. What is y at x = 4?
−7.
3Challenge
Reasoning, generalizing and working backwards.
- ReasoningThe line through (1, a) and (4, 3a) has gradient 4. Find a.
(3a − a) ÷ 3 = 4, so 2a = 12 and a = 6. - ReasoningPoints P(0, 1), Q(2, 5) and R(5, k) are on a straight line. Find k.
Gradient 2; R is 3 across from Q, so k = 5 + 6 = 11.
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? Plot the points on squared paper and draw the triangle by hand.