Mathematics • Year 10 • Unit 2 • Lesson 15
Distance and Midpoint — Mixed Challenge
Six mixed coordinate-geometry problems testing distance, midpoint, and "go backwards" endpoint-finding. Spot a missing-bracket error on a negative coordinate, then design your own isosceles triangle on a grid.
1. Mixed problems — choose distance or midpoint
Show every step including substituting into the formula. Keep exact surds where possible. 3 marks each
1.1 Distance between (−2, 5) and (4, −3).
1.2 Midpoint of (−6, −4) and (10, 2).
1.3 If M(2, −1) is the midpoint of AB and A = (−3, 4), find B.
1.4 Two points have a distance of 13 units. One is at (1, 5). The other has y = 5 (same horizontal line). What are the two possible x-values of the other point?
1.5 Show that the points P(1, 2), Q(4, 6) and R(8, 3) form a right-angled triangle. (Hint: find all three side lengths and use the converse of Pythagoras.)
1.6 A circle has its centre at the midpoint of A(−2, 1) and B(6, 7). Find (a) the centre, (b) the radius (= half the diameter AB), and (c) the area of the circle (use π).
2. Find the mistake
A Year 10 student is finding the distance between (−2, 3) and (4, −1). Their working is below. One line contains a Year-10-realistic sign-or-bracket mistake. Find it. 3 marks
Student's working:
Line 1: d = √((4 − (−2))² + (−1 − 3)²)
Line 2: d = √((4 − 2)² + (−1 − 3)²)
Line 3: d = √(4 + 16)
Line 4: d = √20 = 2√5
(a) Which line is wrong?
(b) Explain the bracket / sign mistake in one or two sentences.
(c) Show the corrected working with the correct exact distance.
Stuck? 4 − (−2) = 4 + 2 = 6, NOT 4 − 2 = 2. The brackets around the negative are essential.3. Open-ended challenge — design your own triangle
Many valid answers. 4 marks
3.1 Design an isosceles triangle on the coordinate plane that has (i) at least one vertex with a negative coordinate, (ii) integer coordinates for all three vertices, (iii) two equal sides each of length exactly 5 units.
Submit:
(a) The three vertex coordinates.
(b) The distance calculation for each of the three sides, showing two of them equal 5.
(c) The midpoint of the unequal side (called the "base midpoint").
How did this worksheet feel?
What I'll revisit before next class:
1.1 — (−2,5) to (4,−3)
Δx = 6, Δy = −8. d = √(36 + 64) = √100 = 10 units.
1.2 — Midpoint (−6,−4) and (10,2)
M = ((−6+10)/2, (−4+2)/2) = (2, −1).
1.3 — Find B given M(2,−1) and A(−3,4)
x_B = 2(2) − (−3) = 7; y_B = 2(−1) − 4 = −6. B = (7, −6). Check midpoint: ((−3+7)/2, (4−6)/2) = (2, −1) ✓.
1.4 — Two possible x-values
Same y so the segment is horizontal: distance = |x₂ − 1| = 13 → x₂ − 1 = ±13 → x₂ = 14 or x₂ = −12.
1.5 — Right-angled triangle
PQ = √(9 + 16) = √25 = 5. QR = √(16 + 9) = √25 = 5. PR = √(49 + 1) = √50. Pythagoras check: PQ² + QR² = 25 + 25 = 50 = PR² ✓. So angle Q is the right angle. The triangle is right-angled at Q.
1.6 — Circle from a diameter
(a) Centre M = ((−2+6)/2, (1+7)/2) = (2, 4). (b) Diameter AB = √((6−(−2))² + (7−1)²) = √(64 + 36) = √100 = 10. Radius = 5 units. (c) Area = π(5)² = 25π ≈ 78.5 square units.
2 — Find the mistake
(a) Line 2.
(b) The student lost the bracket around the negative: 4 − (−2) is +6, not 4 − 2 = 2. Without the bracket the negative gets converted to a subtraction and changes the run from 6 to 2.
(c) Corrected: d = √(6² + (−4)²) = √(36 + 16) = √52 = 2√13 ≈ 7.21 units.
3 — Open-ended (sample solution)
Sample triangle: A(−3, 0), B(3, 0), C(0, 4).
Side lengths: AB = √((3−(−3))² + 0²) = 6 (base, unequal). AC = √(9 + 16) = √25 = 5 ✓. BC = √(9 + 16) = √25 = 5 ✓.
Base midpoint (of AB): ((−3+3)/2, 0) = (0, 0).
Criteria met: A has a negative x-coordinate ✓; all integer vertices ✓; AC = BC = 5 ✓.
Marking: 1 for valid integer vertices including a negative; 1 for correct distance calculations; 1 for two equal sides of length 5; 1 for the base midpoint.