At a glance. One lesson of about 60 minutes. Students write the equation from a gradient and a point (point-slope form), from two points (gradient, then substitution), convert to y = mx + c, and check with the second point.
Calculator: no calculator. Nothing here needs one. Every gradient is a whole number or a simple fraction and the substitutions are short.
What students will learn
Know
- Point-slope form: y − y1 = m(x − x1).
- c = y1 − m x1, found by substitution.
- Two points give the gradient; either point then gives c, and the other checks it.
Understand
- The point-slope form is an equation of the line already; y = mx + c is a tidier version.
- c is the value at x = 0, which is not the y of the point unless the point is on the y-axis.
- The same line has many written forms, and they are equivalent.
Do
- Write point-slope form from a point and a gradient.
- Find the equation through two points and check it.
- Convert point-slope form to y = mx + c.
MYP assessment criteria
Wording paraphrased. What this lesson gives you evidence for, and what to look for:
- Criterion A. Knowing and understanding: finding c and writing the equation (questions 1 to 9).
- Criterion B. Investigating patterns: the first figure. Ask students to find, by changing P with m fixed, what stays the same about the lines, and to write a general statement.
- Criterion C. Communicating: setting out each step of finding an equation clearly, with a check (the worked examples and question 7).
- Criterion D. Applying mathematics in real-life contexts: charging a battery, a falling tank and a retreating glacier (the hook and question 6).
The one thing to do with the figure
Fix m and move P. Ask students to predict c each time, then show it. Many will say c equals P's y; the figure shows c is where the line meets the y-axis, which is only P's y when P is on it. The last state in the steps list (a point on the y-axis) is the one case where they agree.
On the second figure ask for the gradient and c before showing the equation, and use the check with the second point every time.
A possible order
| 5 min | Warm-up from the student page, in silence, then answers. |
|---|---|
| 5 min | Hook: the solar lamp. Collect estimates of the full time and hold them. |
| 10 min | First figure: fix m, move P, predict c. |
| 10 min | Second figure: two points, gradient first. |
| 10 min | Key idea; the worked example settles the hook; students finish the tank. |
| 5 min | Question 1, the check. |
| 10 min | Your turn, Levels 1 to 3. |
| 5 min | Exit ticket, question 9. |
The answers
| Warm-up | Gradient 3 and y-intercept −4; 2; 13. |
|---|---|
| Worked example | y = 10x + 20; full after 8 hours. |
| Faded example | y = −20x + 190. |
| 1. Equation from a point and a gradient | B. A uses y as c; C has a sign slip; D uses the x of the point as the gradient. |
| 2. c from a point and a gradient | 5. y = 2x + 5. |
| 3. Using point-slope form | 7. y − 3 = 2 × 2 = 4. |
| 4. c from two points | 5. y = 2x + 5. |
| 5. c when the gradient is known | 1. y = 3x + 1. |
| 6. Starting value from two readings | 970. The rule is y = −15x + 970. |
| 7. Where a line crosses the y-axis | 7. y = −3x + 7. |
| 8. A gradient from a point and the intercept | 4.5. Rise 9 over run 2. |
| 9. Equation through two points | B. A has a sign slip in c; C uses y as c; D uses the rise as the gradient. |
What each wrong answer tells you
| They give | What it usually means | What to do |
|---|---|---|
| A in question 1; 7 in question 2; 1 in question 4; 4 in question 7 | Used the y of a point as c. | Ask where on the grid c is, and how far the point is from the y-axis. Draw the steps back to x = 0. |
| C in question 1; −5 in question 2 | Sign slip when substituting. | Ask for the check: put the point back in the equation. |
| 4, 15 or 1 in question 3 | Did not use the bracket (x − x1) properly. | Work the bracket on its own first, then multiply, then add. |
| 2 or −3 in question 4 | Stopped at the gradient, or dropped a negative x. | Use brackets when the x is negative: 2 × (−2). |
| −5 or −2 in question 5 | Subtracted the 3x term, or gave the point's y. | Write the substitution line by line and show the sign of 3 × (−1). |
| −15, 910 or 955 in question 6 | Gave the rate, went forward in time, or went back one year. | Draw a time line with years 0, 2 and 6 and the lengths on it. |
| −3 or 1 in question 7 | Gave the gradient, or added the gradient to y. | Ask: from x = 1 to x = 0, does the line go up or down? |
| 9, 0.5 or 2.5 in question 8 | Left out the run, or lost the sign of c. | Ask for the second point on the line, (0, −4), and its rise and run to (2, 5). |
Questions worth asking
- Does the equation change if I choose the other point? No: the line is the same, so both give the same y = mx + c. Listen for “it is one line”.
- Which point is the best to put in? Any works; a convenient one (small whole numbers, or on the y-axis) saves work.
- How can you check your equation? Put the other point in. Listen for “the second point has to fit”.
- What if the two points have the same x? The line is vertical, x = k, with no gradient, so no y = mx + c.
Differentiation
A way in for anyone stuck
- Provide a template: m = …, then y = m x + c, then put a point in, then c = …, then the equation.
- Use points on the y-axis first, where c can be read, then a point one step from it.
- Check each answer by putting both points in the finished equation.
Stretch
- Question 8, then: find the line through (2, 5) that has the same x-intercept as y = 3x − 6. (The x-intercept is (2, 0), so the line is x = 2.)
- Show that the point-slope form through (x1, y1) and through (x2, y2) give the same equation, using the gradient.
Students learning in English as an additional language
“Substitute” means “put in the place of”. Say “put the point in the equation” with a finger on x and then on y. “Point-slope” names the two things you start with.
| Word | Plain meaning | Example |
|---|---|---|
| point-slope form | an equation written from a point and a gradient (also called point-gradient form) | y − 3 = 2(x − 1) |
| substitute | put a number in place of a letter | put x = 2 in y = 4x + c |
| gradient-intercept form | the form y = mx + c | y = 2x + 1 |
| expand | multiply out a bracket | 2(x − 1) = 2x − 2 |
| equivalent | different in form but the same line | y − 3 = 2(x − 1) and y = 2x + 1 |
| check | test an answer with the other point | put the second point back in |
Sentence starters for explaining aloud or in writing:
- The gradient is … so the equation starts y = …x + c.
- I put the point (…, …) in and got c = …
- I checked with the other point and …
Two things not to say
“Use y = mx + c and plug in.” Students need to know what they are finding, c, and why.
“The point is on the line, so c is its y.” Only when the point is on the y-axis.
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.
- c from a pointGradient 3, passes through (2, 10). Find c.
4. - Point-slopeWrite the line with gradient −2 through (1, 5) as y = mx + c.
y = −2x + 7.
2In context
The same idea inside a situation, where the first job is deciding what to calculate.
- ContextA candle is 30 cm tall after 2 hours and 24 cm after 5 hours of burning, at a steady rate. How tall was it at the start?
Rate −2 cm per hour, so 34 cm. - Earlier work (Lesson 3)Where does y = 2x − 6 cross the x-axis?
x = 3.
3Challenge
Reasoning, generalizing and working backwards.
- ReasoningThe line through (1, 2) and (5, k) has the y-intercept 1. Find k.
Gradient 1, so k = 6. - ReasoningFind the equation of the line through (−4, 5) and (2, −1).
Gradient −1, c = 1: y = −x + 1.
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 two points on squared paper and count the steps back to x = 0.