Mathematics • Year 8 • Unit 2 • Lesson 9
Gradient Formula — Mixed Challenge
Pull together everything from Lesson 9: applying the formula, handling negatives, simplifying, recognising collinearity and the special zero/undefined cases. Six mixed problems, one "find the mistake", and one open-ended challenge.
1. Mixed problems — choose the right move
Each question uses a different combination of ideas from Lesson 9. Show all working. 3 marks each
1.1 Find m through (2, 5) and (8, 17).
1.2 Find m through (−4, 3) and (2, −9). Give your answer as a simplified fraction and decimal.
1.3 Find m through (1, 4) and (1, 9). State the gradient type.
1.4 Find m through (−2, 5) and (4, 5). State the gradient type.
1.5 Three points — (1, 4), (3, 10), (6, 19) — are claimed to lie on one straight line. Use the gradient formula on two different pairs to test this claim.
1.6 A line passes through (2, k) and (5, 11). The gradient is m = 2. Find the value of k.
2. Find the mistake
Another student tries to find the gradient through (−1, 2) and (3, −4). Exactly one step is wrong. Spot it, explain why, then re-do correctly. 3 marks
Student's working:
Line 1: Let (x₁, y₁) = (−1, 2) and (x₂, y₂) = (3, −4).
Line 2: m = (y₂ − y₁) / (x₂ − x₁).
Line 3: m = (−4 − 2) / (3 − 1) = −6 / 2 = −3.
Line 4: So m = −3.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write the corrected working and the correct gradient.
Stuck? Subtracting a negative: x₂ − x₁ = 3 − (−1) = 3 + 1 = 4. The student treated −1 as just 1.3. Open-ended challenge — design points with a target gradient
This question has many valid answers. 4 marks
3.1 Find three different pairs of points (each pair distinct from the others) where every pair gives a gradient of m = −2/3.
(a) List your three pairs.
(b) For each pair, show the formula check m = (y₂ − y₁)/(x₂ − x₁) = −2/3.
(c) One sentence: do all three pairs lie on the same line? Why or why not? (Hint: only if at least one point is shared, OR if you check that all 6 points are collinear.)
How did this worksheet feel?
What I'll revisit before next class:
1.1 — (2, 5) and (8, 17)
m = (17 − 5)/(8 − 2) = 12/6 = 2.
1.2 — (−4, 3) and (2, −9)
m = (−9 − 3)/(2 − (−4)) = −12/6 = −2 (= −2.0).
1.3 — (1, 4) and (1, 9)
m = (9 − 4)/(1 − 1) = 5/0 — division by zero. Gradient is undefined (vertical line, x = 1).
1.4 — (−2, 5) and (4, 5)
m = (5 − 5)/(4 − (−2)) = 0/6 = 0. Gradient type: zero (horizontal line, y = 5).
1.5 — Collinearity test
(1,4) → (3,10): m = (10−4)/(3−1) = 6/2 = 3.
(3,10) → (6,19): m = (19−10)/(6−3) = 9/3 = 3.
Both gradients equal 3, so the three points are collinear (they lie on the line y = 3x + 1).
1.6 — Find k
m = (11 − k)/(5 − 2) = 2. So (11 − k)/3 = 2, giving 11 − k = 6, hence k = 5.
2 — Find the mistake
(a) The mistake is on Line 3 (the run calculation).
(b) The student computed x₂ − x₁ as 3 − 1, but x₁ is −1, not 1. Subtracting a negative adds, so it should be 3 − (−1) = 3 + 1 = 4. They lost the sign on x₁.
(c) Corrected: m = (−4 − 2)/(3 − (−1)) = −6/4 = −3/2 = −1.5.
3 — Pairs giving m = −2/3 (sample solution)
Many valid sets. One example:
Pair 1: (0, 0) and (3, −2). m = (−2 − 0)/(3 − 0) = −2/3. ✓
Pair 2: (3, −2) and (6, −4). m = (−4 − (−2))/(6 − 3) = −2/3. ✓
Pair 3: (1, 5) and (4, 3). m = (3 − 5)/(4 − 1) = −2/3. ✓
(c) Pairs 1 and 2 share the point (3, −2), so they lie on the same line: y = (−2/3)x. Pair 3 does NOT lie on that same line (e.g. (1, 5) is not on y = (−2/3)x because (−2/3)(1) = −2/3 ≠ 5). So all three pairs each define a line of gradient −2/3, but they are not all the same line — there are infinitely many parallel lines with the same gradient.
Marking: 1 mark for each valid distinct pair (up to 3 marks) with correct gradient check. 1 mark for a sensible sentence about whether the pairs lie on a single line (parallel-lines reasoning).