Mathematics • Year 8 • Unit 2 • Lesson 12
Equation from a Graph — Mixed Challenge
Pull together everything from Lesson 12: reading c, calculating m from two points, handling negative gradients, lines through the origin, and verifying. Six mixed problems, one "find the mistake", one open-ended challenge.
1. Mixed problems
Show your working: state c, calculate m, write the equation, verify. 3 marks each
1.1 Find the equation of the line through (0, −1) and (4, 7).
1.2 Find the equation of the line through (0, 2) and (5, −3).
1.3 Find the equation of the line through the origin and (4, 10). (Hint: write m as a fraction in simplest form.)
1.4 Find the equation of the line that passes through (−1, 0) and (0, 2). (Read c straight off the second point.)
1.5 A line has gradient 3 and passes through (2, 7). Find c using y = mx + c, then write the equation.
1.6 A line passes through (1, 5) and (3, 11). Find m, then find c, then verify with both points.
2. Find the mistake
A student tried to find the equation of a line through (0, 3) and (2, 9). Their working is shown below. Exactly one line contains a mistake. Spot it, explain why, and re-do correctly. 3 marks
Student's working — line through (0, 3) and (2, 9):
Line 1: c = 3 (from (0, 3)).
Line 2: m = (9 − 3) / (2 − 0) = 6 / 2 = 3.
Line 3: y = mx + c, so y = 3x + 3.
Line 4: Verify at x = 2: y = 3(2) + 3 = 6 + 3 = 12. ✓
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected verification step and confirm the equation is correct.
Stuck? Their equation y = 3x + 3 is fine. So is the m calculation. Verify at (2, 9) — what do you actually get?3. Open-ended challenge — three lines through (0, 4)
This question has more than one valid answer. 4 marks
3.1 Design three different straight lines in y = mx + c form that all cross the y-axis at the same point (0, 4).
For each line you find:
(i) Choose a gradient m (positive, negative, or zero — try a mix).
(ii) Write down the equation.
(iii) Give one extra point on the line by substituting any x-value you like.
Bonus: What do all three lines have in common when drawn on the same set of axes? Describe the picture in one sentence.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — (0, −1) and (4, 7)
c = −1. m = (7 − (−1))/(4 − 0) = 8/4 = 2. y = 2x − 1. Check at (4, 7): 2(4) − 1 = 7 ✓.
1.2 — (0, 2) and (5, −3)
c = 2. m = (−3 − 2)/(5 − 0) = −5/5 = −1. y = −x + 2. Check at (5, −3): −(5) + 2 = −3 ✓.
1.3 — Origin and (4, 10)
c = 0. m = 10/4 = 5/2 (or 2.5). y = (5/2)x or y = 2.5x. Check at (4, 10): (5/2)(4) = 10 ✓.
1.4 — (−1, 0) and (0, 2)
c = 2 (from the second point). m = (2 − 0)/(0 − (−1)) = 2/1 = 2. y = 2x + 2. Check at (−1, 0): 2(−1) + 2 = 0 ✓.
1.5 — Gradient 3, through (2, 7)
Substitute into y = mx + c: 7 = 3(2) + c → 7 = 6 + c → c = 1. y = 3x + 1.
1.6 — Through (1, 5) and (3, 11)
m = (11 − 5)/(3 − 1) = 6/2 = 3. Use (1, 5): 5 = 3(1) + c → c = 2. y = 3x + 2. Verify: (1, 5) → 3 + 2 = 5 ✓; (3, 11) → 9 + 2 = 11 ✓.
2 — Find the mistake
(a) The mistake is in Line 4 (the verification).
(b) Lines 1–3 are all correct: c = 3, m = 3, equation y = 3x + 3. But verifying at x = 2 the student wrote y = 12, not y = 9 — yet (2, 9) is the second known point. Actually 3(2) + 3 = 6 + 3 = 9, NOT 12. The student must have added wrong, or used (2, 12) instead of (2, 9).
(c) Correct verification: at x = 2, y = 3(2) + 3 = 6 + 3 = 9. This matches the given point (2, 9), confirming y = 3x + 3 is correct.
3 — Open-ended challenge (sample solution)
All equations have c = 4, so each has the form y = mx + 4.
Line 1 (positive gradient): m = 2 → y = 2x + 4. At x = 1: y = 6, so (1, 6) lies on it.
Line 2 (negative gradient): m = −1 → y = −x + 4. At x = 4: y = 0, so (4, 0) lies on it.
Line 3 (horizontal): m = 0 → y = 4. Every point has y = 4, e.g. (10, 4).
Bonus: All three lines pass through the same point (0, 4) on the y-axis — they form a "fan" of lines pivoting on that y-intercept.
Marking: 1 mark per valid distinct line (3 marks total). 1 bonus mark for the clear description that all three meet at (0, 4).