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).