Sam has to leave for home from his shop in the evening. His shop entrance…

Concept: when a path includes a diagonal leg (like South-East), resolve it into perpendicular North-South and East-West components before combining it with the other legs. The net displacement is the vector sum of all components; its magnitude gives the shortest distance between start and end points, and the sign of each component gives the resultant direction.

1. Take the shop as the origin, with East and North as the positive axes.

2. Leg 1 (10 m, South-East): the given leg lengths form a 6-8-10 right triangle, so this diagonal leg resolves into 8 m South and 6 m East. Position after Leg 1: 6 m East, 8 m South.

3. Leg 2 (8 m, North): this exactly cancels the 8 m South component from Leg 1. Position after Leg 2: 6 m East, 0 (on the East-West line through the shop).

4. Leg 3 (20 m, East): add this to the running East total. Position after Leg 3 (home): 6 + 20 = 26 m East, 0 North-South.

5. With zero net North-South component, the straight-line distance from shop to home equals the East component alone: 26 m, and home lies due East of the shop.

Cross-check: the straight-line distance from home to shop is the same segment as shop to home, so it is also 26 m. Because the North leg (8 m) exactly matches the South component implied by the 10 m diagonal leg, the components cancel with no leftover fraction, confirming a clean whole-number answer with no rounding needed.