Using vectors and the Pythagorean Theorem to find distances — problems for a 14‑year‑old (Prealgebra, draw a picture)

Step‑by‑step solutions using pictures and the Pythagorean theorem to find distances after a sequence of east/west and north/south moves. Two worked problems with clear coordinates and a simple diagram. CCSS: A‑SSE.1 and A‑CED.1.

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I'll walk you through both problems step by step. For each one we'll place points on a coordinate grid (east = +x, north = +y), draw a simple picture, compute horizontal and vertical net displacements, then use the Pythagorean theorem to get the straight‑line distance.

Problem 1

W swims: 60 miles north, 30 miles east, 30 miles north, 150 miles west. How far is W from the starting point?

  1. Put the start at (0,0). Track x (east) and y (north) separately.
    • After 60 mi north: (0, 60).
    • Then 30 mi east: (30, 60).
    • Then 30 mi north: (30, 90).
    • Then 150 mi west: (30 - 150, 90) = (-120, 90).
  2. Net displacement from start = (-120, 90). That means 120 miles west and 90 miles north.
  3. Distance from start = sqrt((120)^2 + (90)^2).
    Compute: 120^2 = 14400, 90^2 = 8100, sum = 22500.
    sqrt(22500) = 150.

Answer (Problem 1): 150 miles from the starting point.

Problem 2

Points and relations given: A is 50 m east of B and 30 m west of C. D is 60 m east of C and 40 m east of E. F is 50 m north of E and 80 m north of G. Find the distance between B and G (to the nearest tenth).

  1. Choose coordinates: place B at (0,0). East is +x, north is +y.
  2. Use the given east/west relations (these put A, B, C, D, E on the same horizontal line):
    • A is 50 m east of B ⇒ A = (50, 0).
    • A is 30 m west of C ⇒ C = (A_x + 30, 0) = (80, 0).
    • D is 60 m east of C ⇒ D = (C_x + 60, 0) = (140, 0).
    • D is 40 m east of E ⇒ D_x = E_x + 40 ⇒ E_x = D_x - 40 = 100. So E = (100, 0).
  3. Now use the north/south relations. The wording "F is 50 m north of E and 80 m north of G" means F is directly north of E and also directly north of G (so E and G share F's x):
    • F is 50 m north of E ⇒ F = (100, 50).
    • F is 80 m north of G ⇒ G has same x = 100 and y = 50 - 80 = -30. So G = (100, -30).
  4. Now we have B = (0,0) and G = (100, -30). Compute the straight‑line distance:
    Δx = 100 − 0 = 100, Δy = −30 − 0 = −30.
    Distance = sqrt(100^2 + (−30)^2) = sqrt(10000 + 900) = sqrt(10900) ≈ 104.403065... m.
    To the nearest tenth: 104.4 m.
Simple sketch (not to scale):

    y
    ^
  50|                F(100,50)
    |                |
   0| B(0,0) ---- A(50,0) ---- C(80,0) ---- E(100,0) ---- D(140,0)
    |
 -30|                G(100,-30)
    +---------------------------------> x

Answer (Problem 2): B and G are about 104.4 meters apart (to the nearest tenth).

Notes connecting to standards: these solutions use seeing structure in movements (A‑SSE.1) by separating horizontal and vertical components and translating the description into coordinates/equations (A‑CED.1). The Pythagorean theorem gives the straight‑line distance from the component distances.

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