Show that A(−1, 0), B(3, 1), C(2, 5) and D(−2, 4) are the vertices of a square.

1. 1 · Write what is given and what is to be found. Given: A(−1, 0), B(3, 1), C(2, 5), D(−2, 4). To show: ABCD is a square.
2. 2 · Draw the figure. Draw a rough quadrilateral ABCD with the vertices in order, and draw both diagonals AC and BD.
3. 3 · Write the formula or theorem. Distance formula: d = √[(x₂ − x₁)² + (y₂ − y₁)²]
4. 4 · Find the four sides.
   • AB = √(4² + 1²) = √17
   • BC = √[(−1)² + 4²] = √17
   • CD = √[(−4)² + (−1)²] = √17
   • DA = √[1² + (−4)²] = √17
5. 5 · Find the two diagonals.
   • AC = √(3² + 5²) = √34
   • BD = √[(−5)² + 3²] = √34
6. 6 · Conclude. All four sides are equal and the diagonals are equal. Therefore ABCD is a square.
7. 7 · Check and box the answer. Check: AB² + BC² = 17 + 17 = 34 = AC², so ∠B = 90° (converse of Pythagoras' theorem). ABCD is a square.

Scoring tip: Equal sides alone prove only a rhombus. Always add the diagonals, or a right angle, to prove a square.