Beast Academy 5D (Age 15) — Pythagorean Paths: Find Distance Between Two Points (8.G.B.8) — Practice Help for 5D pg.58–60 (#142–155)

Step-by-step guide for a 15-year-old to apply the Pythagorean Theorem and the distance formula to find distances between two points on a coordinate plane. Worked examples, tips, common mistakes, and practice strategy for Beast Academy 5D pages 58–60 (problems #142–155, #151–155 optional).

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Goal: Use the Pythagorean Theorem to find the distance between two points (x1,y1) and (x2,y2) on a coordinate plane.

Quick idea: The horizontal difference and vertical difference between the points are the two legs of a right triangle. The distance is the hypotenuse.

  1. Plot or note the points. Write the coordinates clearly: (x1, y1) and (x2, y2).
  2. Find the horizontal change (dx) and vertical change (dy).

    dx = x2 − x1, dy = y2 − y1. (You can use absolute values for the lengths: |dx| and |dy|.)

  3. Build the right triangle. One leg = |dx|, the other leg = |dy|. The distance d is the hypotenuse.
  4. Apply the Pythagorean Theorem.

    d = sqrt((dx)^2 + (dy)^2) = sqrt((x2 − x1)^2 + (y2 − y1)^2).

  5. Simplify. If the square root simplifies to an integer or a simpler radical, do that. Otherwise leave as a simplified radical or decimal.

Derivation (why that formula works):

If dx = x2 − x1 and dy = y2 − y1, then by Pythagorean theorem, d^2 = (dx)^2 + (dy)^2, so d = sqrt((x2 − x1)^2 + (y2 − y1)^2).

Worked examples

Example 1: Find the distance between (1, 2) and (5, 6).

  • dx = 5 − 1 = 4, dy = 6 − 2 = 4.
  • d = sqrt(4^2 + 4^2) = sqrt(16 + 16) = sqrt(32) = 4√2.

Example 2: Find the distance between (−3, 4) and (3, −2).

  • dx = 3 − (−3) = 6, dy = −2 − 4 = −6 (length = 6).
  • d = sqrt(6^2 + (−6)^2) = sqrt(36 + 36) = sqrt(72) = 6√2.

Example 3: Between (0,0) and (1,2).

  • dx = 1, dy = 2 → d = sqrt(1 + 4) = sqrt(5).

Tips for Beast Academy practice (pg. 58–60, #142–155):

  • Always write dx and dy before squaring — it reduces sign errors.
  • If points lie on the same horizontal or vertical line, distance is just |dx| or |dy| (no radical needed).
  • Recognize Pythagorean triples (3-4-5, 5-12-13, etc.) so you can simplify quickly to integers when possible.
  • When you get a radical like sqrt(72), factor perfect squares: sqrt(72)=sqrt(36·2)=6√2.
  • Order of subtraction doesn't matter because you square the result: (x2−x1)^2 = (x1−x2)^2.
  • For optional/problems #151–155, expect slightly trickier setups — practice drawing the right triangle carefully and labeling leg lengths before computing.

Common mistakes to avoid

  • Forgetting to square both dx and dy.
  • Mixing up which coordinate to subtract from which — but remember squaring removes the sign worry.
  • Failing to simplify radicals (leave them simplified: factor out perfect squares).
  • Not drawing a quick sketch — a diagram helps spot right triangles and Pythagorean triples.

Practice checklist before submitting answers:

  • Did I compute dx and dy correctly?
  • Did I square each and add them before taking sqrt?
  • Is my final answer simplified (integer, simplified radical, or correct decimal)?
  • Does the answer seem reasonable on the drawing (longer distance for farther-apart points)?

If you want, tell me one specific problem number from #142–155 and I will walk through that exact problem step-by-step with a diagram and full simplification.

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