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Goal

Use the Pythagorean Theorem to find the straight-line distance between two points on a dot-grid (coordinate plane). We'll make four diagrams where the distances are: √10, √10, 5, and √10.

Step-by-step method (what to do every time)

  1. Write the coordinates of the two points: (x1, y1) and (x2, y2).
  2. Find the horizontal difference dx = |x2 − x1| and the vertical difference dy = |y2 − y1|.
  3. Use the Pythagorean Theorem: distance = sqrt(dx^2 + dy^2). This works because dx and dy are the legs of a right triangle and the distance is the hypotenuse.
  4. Simplify the square root if possible (for example, sqrt(25) = 5).

Diagrams (dot-grid) with calculations

Each diagram below shows a small dot-grid (x and y are integers). The bold points are the two points whose distance we find. The right triangle is drawn so you can see dx and dy.

Diagram 1 — distance = √10
(0,0) (1,3) dx = 1 dy = 3

Calculation: dx = |1−0| = 1, dy = |3−0| = 3. Distance = √(1^2 + 3^2) = √(1 + 9) = √10.

Diagram 2 — distance = √10 (different orientation)
(0,0) (3,1) dx = 3 dy = 1

Calculation: dx = |3−0| = 3, dy = |1−0| = 1. Distance = √(3^2 + 1^2) = √(9 + 1) = √10.

Diagram 3 — distance = 5 (a 3-4-5 triangle)
(0,0) (3,4) dx = 3 dy = 4

Calculation: dx = |3−0| = 3, dy = |4−0| = 4. Distance = √(3^2 + 4^2) = √(9 + 16) = √25 = 5.

Diagram 4 — distance = √10 (shifted triangle)
(2,2) (3,5) dx = 1 dy = 3

Calculation: dx = |3−2| = 1, dy = |5−2| = 3. Distance = √(1^2 + 3^2) = √(1 + 9) = √10.


Quick tips:

  • If dx and dy are integers, you often get familiar square-root results (like 5 from a 3-4-5 triangle).
  • Look for small integer dx,dy pairs that satisfy dx^2 + dy^2 = N when you want to make a distance √N. For √10, try (1,3) or (3,1).
  • Always label the coordinates and check dx, dy carefully — sign doesn't matter because you take absolute differences.

Related standard: 8.G.B.8 — applying the Pythagorean Theorem to find distances in the coordinate plane.


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