Have you ever wondered if geometry could be a little like dating in the city — one must choose the right partner, in the right order, to make the evening perfect? That’s the mood of our little grid romance. We have five dots: A=(5,0), B=(6,3), C=(1,3), D=(6,2), E=(3,2). Our mission — like an exacting dinner reservation — is to line them up so the four consecutive distances are precisely 1, √10, 5, √5, in that order.
First, whisper to yourself the Pythagorean rule: distance squared = (Δx)² + (Δy)². It’s the recipe card. Now, like sniffing a perfume to find the right note, look for the unique tiny scent on the grid: which pair of points is exactly 1 apart? Checking the pairs, you’ll quickly spot that B(6,3) and D(6,2) differ by (0,1), so their squared distance is 0² + 1² = 1. That gives us our opening move — the only possible adjacency for the required length 1. If a dance must start with a whisper, this is it.
From that forced start, we try to follow the list. Put the points in letters and attempt the chain D → B → ? and see what fits next. Compute the next distances using simple subtraction and squaring — tasting each candidate so you don’t miss the note:
D(6,2) → B(6,3): dx=0, dy=1 ⇒ 0²+1² = 1 (distance = 1) B(6,3) → A(5,0): dx=-1, dy=-3 ⇒ 1+9 = 10 (distance = √10) A(5,0) → C(1,3): dx=-4, dy=3 ⇒ 16+9 = 25 (distance = 5) C(1,3) → E(3,2): dx=2, dy=-1 ⇒ 4+1 = 5 (distance = √5)
So the path D → B → A → C → E (that is, (6,2) → (6,3) → (5,0) → (1,3) → (3,2)) gives exactly the required sequence: 1, √10, 5, √5. Like finding the perfect outfit, once the first accessory (the unique length-1 pair) is chosen, the rest falls into place gracefully.
Why unique? Because that lonely distance of 1 only appears once among all pairs, and adjacency is required. Once you pin that down, the next distances become strong hints — √10 is rarer on our grid than, say, √5, so the path is pruned quickly until only the single continuous chain works without reusing points.
Step-by-step strategy I teach to students: 1) Label the points, 2) Make a table of pairwise squared distances (avoid roots until the end), 3) Find any unique forced adjacency (like our 1), 4) Extend step-by-step matching the required order, and 5) Verify each square-sum equals the needed squared lengths (1, 10, 25, 5).
Finally, a little assessment nibble: Proficient work shows the correct path, clear calculations using Pythagoras, and readable steps. Exemplar work lays out *every* pair checked, explains why other orders fail, and includes a neat labelled diagram. In short — be systematic, show your arithmetic, and savour the tidy satisfaction of a path that clicks. Who knew grids could be so chic?