Mathematics • Year 10 • Unit 2 • Lesson 16
Gradient — Mixed Challenge
Six mixed gradient problems testing two-point gradients, rearranging general form, classifying horizontal/vertical/diagonal lines, and finding missing coordinates. Spot a swapped rise/run, then design a line that satisfies three conditions.
1. Mixed problems — choose the right method
Show the formula or the rearrangement, then the result. 3 marks each
1.1 Find the gradient through (−4, 6) and (8, 0).
1.2 Find the gradient of 2x − 5y + 10 = 0.
1.3 A line has gradient m = 3 and passes through (2, k) and (5, 10). Find k.
1.4 Find the gradient of the line through (−1, −3) and (4, −13). Simplify the fraction and state the sign.
1.5 A line of gradient 2/3 passes through (3, 4). What is the y-coordinate when x = 9? (Hint: use rise = m × run.)
1.6 Classify each line by its gradient: (a) y = 7 (a horizontal line), (b) x = −2 (a vertical line), (c) y = −x + 5, (d) the line through (1, 1) and (3, 1).
2. Find the mistake
A Year 10 student calculates the gradient through A(2, 7) and B(5, 1). Their working is below. One line contains a swap-the-rise-and-run mistake. Find it. 3 marks
Student's working:
Line 1: (x₁, y₁) = (2, 7); (x₂, y₂) = (5, 1)
Line 2: m = (x₂ − x₁)/(y₂ − y₁)
Line 3: m = (5 − 2)/(1 − 7)
Line 4: m = 3/−6 = −1/2
(a) Which line is wrong?
(b) Explain the formula mix-up in one sentence.
(c) Show the corrected working with the correct gradient.
Stuck? Gradient is RISE OVER RUN — that is (y₂ − y₁) on top, (x₂ − x₁) on the bottom. The student inverted the formula.3. Open-ended challenge — design a line
Many valid answers. 4 marks
3.1 Design a line that satisfies all three of: (i) gradient is exactly −2/3, (ii) the line passes through a point with a negative y-coordinate, (iii) the line passes through the point (3, 0).
Submit:
(a) Two points on the line (one must satisfy (ii) and one must be (3, 0)).
(b) The gradient calculation showing it equals −2/3.
(c) The equation of the line in y = mx + c form.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — (−4,6) to (8,0)
m = (0 − 6)/(8 − (−4)) = −6/12 = −1/2.
1.2 — 2x − 5y + 10 = 0
−5y = −2x − 10 → y = (2/5)x + 2. m = 2/5.
1.3 — m = 3 through (2,k) and (5,10)
3 = (10 − k)/(5 − 2) → 10 − k = 9 → k = 1.
1.4 — (−1,−3) to (4,−13)
m = (−13 − (−3))/(4 − (−1)) = −10/5 = −2 (negative — line slopes downward).
1.5 — m = 2/3 from (3,4) to x = 9
run = 9 − 3 = 6; rise = (2/3)(6) = 4. So y = 4 + 4 = 8. The point is (9, 8).
1.6 — Classify
(a) y = 7: horizontal, m = 0. (b) x = −2: vertical, m = undefined. (c) y = −x + 5: m = −1, downward slope. (d) Through (1,1) and (3,1): both y = 1, horizontal, m = 0.
2 — Find the mistake
(a) Line 2.
(b) The student wrote run/rise instead of rise/run — the formula is m = (y₂ − y₁)/(x₂ − x₁), not the other way around.
(c) Corrected: m = (1 − 7)/(5 − 2) = −6/3 = −2. (The numerical answer happens to be the negative reciprocal of the student's answer — common consequence of inverting the formula.)
3 — Open-ended (sample solution)
Sample two points: (3, 0) and (6, −2) (negative y ✓).
Gradient check: m = (−2 − 0)/(6 − 3) = −2/3 ✓.
Equation: y − 0 = (−2/3)(x − 3) → y = (−2/3)x + 2. So y = −(2/3)x + 2.
Verify (3, 0): y = −(2/3)(3) + 2 = −2 + 2 = 0 ✓.
Marking: 1 for two valid points satisfying the conditions; 1 for the gradient calculation; 1 for the y = mx + c equation; 1 for a substitution verifying it passes through (3, 0).