Goal: Use the Pythagorean Theorem to find the distance between two points on a dot grid or coordinate plane.
Key ideas:
- If you have two points (x1, y1) and (x2, y2), the horizontal difference is Δx = x2 − x1 and the vertical difference is Δy = y2 − y1. The distance d between the points is
d = sqrt((Δx)^2 + (Δy)^2). - On a dot grid, you can draw a right triangle where the horizontal and vertical legs are Δx and Δy, and the hypotenuse is the distance between the two points. Then apply the Pythagorean Theorem: leg^2 + leg^2 = hypotenuse^2.
Step-by-step method (use this every time):
- Pick the two points; compute Δx and Δy (count how many dots horizontally and vertically).
- Square Δx and Δy, add the squares: (Δx)^2 + (Δy)^2.
- Take the square root: d = sqrt((Δx)^2 + (Δy)^2).
Worked examples with dot-grid diagrams. Each diagram uses a right triangle built from the origin so the hypotenuse is the distance between (0,0) and the labeled point.
Distance = √10 (example)
Points: (0,0) and (3,1). Δx = 3, Δy = 1. Compute: 3^2 + 1^2 = 9 + 1 = 10, so distance = √10.
Distance = √17 (example)
Points: (0,0) and (4,1). Δx = 4, Δy = 1. Compute: 4^2 + 1^2 = 16 + 1 = 17, so distance = √17.
Distance = √8 (example)
Points: (0,0) and (2,2). Δx = 2, Δy = 2. Compute: 2^2 + 2^2 = 4 + 4 = 8, so distance = √8 = 2√2.
Practice tip (Beast Academy 5D pg. 58-60 #142–155): For each problem, draw the two points on the dot grid, make a right triangle by dropping a vertical and horizontal segment, count Δx and Δy, then compute d = sqrt((Δx)^2 + (Δy)^2). Problems #151–155 are optional — use them to challenge yourself with longer steps or non-origin starting points.
Quick reminder: if Δx and Δy are both integers, the distance is sqrt(integer). Some distances simplify (like √8 = 2√2). For other sums you may leave the answer as a square root.
Ask me a specific problem from the practice set if you want a step-by-step solution for that question.