Pythagorean Paths — Find Distance Between Two Points (20-year-old, Beast Academy 5D, Pythagorean Paths Enrichment)

Step-by-step guide (age 20, Beast Academy 5D) for using the Pythagorean Theorem and the distance formula to find distances between points on a coordinate dot grid. Includes three worked examples, SVG coordinate dot-grid illustrations, tips for Practice 5D pg.58–60 (#142–155) and optional problems #151–155.

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Pythagorean Paths — What you need to know

Goal: Given two points (x1, y1) and (x2, y2) on a coordinate grid, find the straight-line distance between them using the Pythagorean Theorem.

Step-by-step method

  1. Plot the two points on the coordinate grid.
  2. Draw a right triangle by making a horizontal segment and a vertical segment between the points (these are the legs).
  3. Find the lengths of the horizontal and vertical legs: Δx = |x2 - x1| and Δy = |y2 - y1|.
  4. Use the Pythagorean Theorem: distance^2 = (Δx)^2 + (Δy)^2. So distance = sqrt((x2 - x1)^2 + (y2 - y1)^2).
  5. Simplify the square root if possible (look for Pythagorean triples or factor out perfect squares).

The formula you use every time is the distance formula:

distance = √[(x2 - x1)2 + (y2 - y1)2]

Example 1 — Horizontal points (easy)

Points: A = (1, 2), B = (4, 2).

Step 1: Δx = |4 - 1| = 3, Δy = |2 - 2| = 0. The points are on the same horizontal line.

Step 2: distance = √(32 + 02) = √9 = 3.

A(1,2) B(4,2)

Example 2 — A Pythagorean triple (nice and clean)

Points: A = (1, 2), B = (4, 6).

Step 1: Δx = |4 - 1| = 3, Δy = |6 - 2| = 4.

Step 2: distance = √(32 + 42) = √(9 + 16) = √25 = 5.

A(1,2) B(4,6) leg=3 leg=4 hyp=5

Because 3-4-5 is a Pythagorean triple, the distance is exactly 5.

Example 3 — Negative coordinates and a non-perfect square

Points: A = (-2, 1), B = (3, 4).

Step 1: Δx = |3 - (-2)| = 5, Δy = |4 - 1| = 3.

Step 2: distance = √(52 + 32) = √(25 + 9) = √34.

We leave it as √34 (can't simplify to an integer). If you need a decimal, √34 ≈ 5.8309519.

A(-2,1) B(3,4) Δx=5 Δy=3

Tips for Practice (Beast Academy 5D pages 58–60, problems #142–155)

  • Always plot the points first — the right triangle is obvious on the grid.
  • Count squares for Δx and Δy (use absolute differences so order doesn’t matter).
  • Look for Pythagorean triples (3-4-5, 5-12-13, etc.) to get integer distances quickly.
  • If the result is a square root, see if you can factor out perfect squares (e.g., √72 = √(36·2) = 6√2).
  • For optional problems (#151–155), they may include negative coordinates or require simplifying radicals — use the same steps.

Quick practice (try these before checking answers)

  1. Distance between (2, 3) and (7, 3).
  2. Distance between (0, 0) and (3, 4).
  3. Distance between (-1, -2) and (2, 1).

Answers: 1) 5. 2) 5 (3-4-5). 3) Δx=3, Δy=3, distance = √(18) = 3√2.

If you want, I can: show step-by-step solutions to any specific problem from pages 58–60 (#142–155), generate more dot-grid diagrams, or convert any problem into an SVG so you can visualize it exactly. Which problems would you like me to do next?

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