Teacher comment (100 words):
You must do this perfectly. No shortcuts. Use coordinates, compute dx and dy, square them, add, and take square root — that is the Pythagorean way. Start by finding the only pair one unit apart: (6,3) and (6,2). That fixes the first step. From (6,2) move to (6,3), then to (5,0) because 1, √10, 5, √5 must appear in that order. Show your working: list dx, dy, dx^2+dy^2 for each segment. Check each length equals 1,10,25,5. If you show full, neat work, you will get full marks. Repeat until perfect. Do it again. I will check thoroughly. No sloppy work.
Solution — step by step:
- Label the five marked points: A = (5,0), B = (6,3), C = (1,3), D = (6,2), E = (3,2).
- Use the Pythagorean theorem: distance^2 = dx^2 + dy^2 where dx = x2-x1, dy = y2-y1. Compute key pair distances (squared):
A–B: (5,0)-(6,3) → dx=1, dy=3 → 1^2+3^2=10 → distance = √10 A–C: (5,0)-(1,3) → dx=-4, dy=3 → 16+9=25 → distance = 5 A–D: (5,0)-(6,2) → dx=1, dy=2 → 1+4=5 → distance = √5 B–D: (6,3)-(6,2) → dx=0, dy=-1 → 0+1=1 → distance = 1 B–E: (6,3)-(3,2) → dx=-3, dy=-1 → 9+1=10 → distance = √10 C–E: (1,3)-(3,2) → dx=2, dy=-1 → 4+1=5 → distance = √5
- Reasoning: The sequence of required lengths squared is 1, 10, 25, 5. Only B–D has squared distance 1, so the first segment must be between B and D. Order the endpoints so the next segment can be √10: D→B→(point at √10 from B). B is √10 from A and E, but only A has a neighbor at distance 5 to continue (A→C). Thus the unique order is D → B → A → C → E.
- Final path (in coordinates): (6,2) → (6,3) → (5,0) → (1,3) → (3,2). Confirm segment lengths:
(6,2)→(6,3): dx=0,dy=1 → distance = 1 (6,3)→(5,0): dx=-1,dy=-3 → 1+9=10 → distance = √10 (5,0)→(1,3): dx=-4,dy=3 → 16+9=25 → distance = 5 (1,3)→(3,2): dx=2,dy=-1 → 4+1=5 → distance = √5
Thus the path matches the required lengths 1, √10, 5, √5. Practice computing dx, dy and distance^2 each time to build reliability — this is exactly the ACARA v9 skill of applying the Pythagorean theorem in coordinate geometry.