Pythagorean Paths: Using the Pythagorean Theorem to Find Distances Between Points — Age 20 — Beast Academy 5D (8.G.B.8)

Clear, step-by-step explanation of how to apply the Pythagorean Theorem to find the distance between two points in a coordinate plane. Worked examples, formula derivation, tips, and a set of practice problems (required and optional).

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Goal

Find the distance between two points (x1, y1) and (x2, y2) in the coordinate plane by turning the difference in coordinates into the legs of a right triangle and using the Pythagorean Theorem.

Derivation / Distance Formula (step by step)

  1. Plot the two points. Draw a horizontal line from one point and a vertical line from the other so they meet at a right angle — this makes a right triangle.
  2. The horizontal leg length is the absolute difference in x-values: |x2 - x1|. The vertical leg length is the absolute difference in y-values: |y2 - y1|.
  3. By the Pythagorean Theorem, if the legs are a and b and the hypotenuse is d, then d^2 = a^2 + b^2. Here a = x2 - x1 and b = y2 - y1 (signs don’t matter because we square them). Therefore 
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)

This is the distance formula. It is symmetric: swapping the two points gives the same result.

Worked examples

Example 1 (integer result)

Find the distance between A = (1, 2) and B = (4, 6).

  1. Compute differences: x2 - x1 = 4 - 1 = 3; y2 - y1 = 6 - 2 = 4.
  2. Apply the formula: d = sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5.
  3. Interpretation: The points are 5 units apart (a 3-4-5 right triangle).

Example 2 (nontrivial signs and simplification)

Find the distance between P = (−3, 4) and Q = (5, −2).

  1. Differences: x2 - x1 = 5 - (−3) = 8. y2 - y1 = −2 - 4 = −6. (We will square these, so the negative sign won’t matter.)
  2. d = sqrt(8^2 + (−6)^2) = sqrt(64 + 36) = sqrt(100) = 10.
  3. Answer: distance = 10.

Example 3 (irrational distance)

Find the distance between (0,0) and (1,1).

  1. Differences: 1 - 0 = 1 and 1 - 0 = 1.
  2. d = sqrt(1^2 + 1^2) = sqrt(2).
  3. Answer: distance = sqrt(2) (approximately 1.414).

Special cases & tips

  • If x1 = x2, the distance is just |y2 - y1| (vertical segment). If y1 = y2, the distance is |x2 - x1| (horizontal segment).
  • Order doesn’t matter: swapping the points gives the same differences squared.
  • When differences are integers that form a Pythagorean triple (3-4-5, 5-12-13, etc.), the distance is an integer. If not, leave the answer as a simplified radical (e.g., sqrt(32) = 4*sqrt(2)).
  • For fractions or decimals, compute the differences and square them; you may need to rationalize or give a decimal approximation depending on the problem instructions.

Practice problems (Beast Academy style: 5D pg. 58-60 — problems 142-155). Required: 142–150. Optional: 151–155.

  1. 142. Distance between (1,2) and (4,6).
  2. 143. Distance between (−3,4) and (5,−2).
  3. 144. Distance between (0,0) and (1,1).
  4. 145. Distance between (2,5) and (2,−3).
  5. 146. Distance between (7,1) and (3,1).
  6. 147. Distance between (−1,−1) and (2,3).
  7. 148. Distance between (1/2, 3/2) and (−1/2, 3/2).
  8. 149. Distance between (0,5) and (12,0).
  9. 150. Distance between (2,2) and (6,6).

Optional:

  1. 151. Distance between (−2,0) and (1,4).
  2. 152. Distance between (3,−4) and (−1,2).
  3. 153. Distance between (0,0) and (5,12).
  4. 154. Distance between (1.5, 2.5) and (4.5, 6.5).
  5. 155. Distance between (−3/2, 0) and (3/2, 0).

Answers (check your work)

  • 142: 5
  • 143: 10
  • 144: sqrt(2)
  • 145: 8
  • 146: 4
  • 147: 5
  • 148: 1
  • 149: 13
  • 150: sqrt(32) = 4·sqrt(2)
  • 151 (optional): 5
  • 152 (optional): sqrt(52) = 2·sqrt(13)
  • 153 (optional): 13
  • 154 (optional): 5
  • 155 (optional): 3

Final advice

Always write down the differences in x and y first, square them, add them, then take the square root. Draw the right triangle if you feel unsure. For contest work, look for Pythagorean triples to simplify answers quickly.

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