A person travels 12 km due North, then 15 km due East, after that 15 km due…

When a person walks through a sequence of directional legs, the straight-line distance back to the starting point (the resultant displacement) is found by resolving every leg onto two perpendicular axes — North-South and East-West — and combining the legs on each axis. Movements in opposite directions on the same axis cancel each other out (they subtract); if the net displacement on one axis works out to zero, the resultant distance equals just the magnitude of the net displacement on the other axis.

1. Group the four legs onto the two axes: North (12 km) and South (18 km) lie on the North-South axis; East (15 km) and West (15 km) lie on the East-West axis.

2. Net East-West displacement = 15 km East − 15 km West = 0 km, so the East and West legs cancel out exactly.

3. Net North-South displacement = 12 km North − 18 km South = 6 km net towards the South.

4. Since the East-West component is 0 km, the straight-line distance from the starting point equals the magnitude of the North-South component alone, i.e. 6 km.

As a check, using the general resultant-distance formula, distance = √((net East-West)² + (net North-South)²) = √(0² + 6²) = √36 = 6 km, which matches the direct reading above.

So the person ends up 6 km from the starting point.