290. The Distance Formula · measuring straight lines on a grid

Try to break it

Drag A and B anywhere. The horizontal leg |x₂ − x₁| and the vertical leg |y₂ − y₁| change with them, and the distance √((x₂−x₁)² + (y₂−y₁)²) is the hypotenuse of that right triangle. Align A and B on the same x and the formula collapses to |y₂ − y₁|; align them on the same y and it collapses to |x₂ − x₁|.

How you build it

Construct the right triangle whose hypotenuse is the distance from A to B.

• Plot A = (-4, -2): from O count 4 left, then 2 down. Click that grid point.
• Plot B = (4, 2): from O count 4 right, then 2 up. Click that grid point.
• Plot C = (4, -2) — directly below B and level with A. This is the right-angle corner of the triangle.
• Draw the leg from A to C. It runs flat along y = -2, so its length is the x-gap, |x₂ − x₁| = 8.
• Draw the leg from C to B. It runs straight up at x = 4, so its length is the y-gap, |y₂ − y₁| = 4.
• Draw the hypotenuse from A to B. By Pythagoras, AB² = AC² + BC², so d = √(8² + 4²) = √80 ≈ 8.9 — the distance formula.

The proof, step by step

Prove that the distance between two points is √((x₂−x₁)² + (y₂−y₁)²).

• Identify the horizontal leg AC = |x₂ – x₁|.
• Identify the vertical leg BC = |y₂ – y₁|.
• Apply Pythagoras' Theorem: AB² = AC² + BC².
• Conclude AB = √((x₂ – x₁)² + (y₂ – y₁)²).

Worked example

In a Cartesian plane, point P is at (2, 3) and point Q is at (8, 11). What is the distance between P and Q?

Using the distance formula: PQ = √((8–2)² + (11–3)²) = √(36 + 64) = √100 = 10 units.

• 8 units
• 10 units — correct
• 12 units
• 14 units