Two girls A and B start from the same point. A walks 9 m North, then turns to…

Direction-and-distance problems are solved on a coordinate plane: place the shared starting point at the origin, take North as +y and East as +x, and treat each turn as a 90° change of facing (a right turn is clockwise, a left turn is counter-clockwise). A walker's final position is the vector sum of their displacements, and one walker's position relative to another is simply the difference between their final coordinates.

1. Place the shared starting point at the origin (0, 0), with North along +y and East along +x.

2. Girl A walks 9 m North, moving from (0, 0) to (0, 9).

3. Facing North, A's right turn changes her facing to East; walking 5 m East takes her from (0, 9) to (5, 9) — A's final position.

4. Girl B walks 11 m East, moving from (0, 0) to (11, 0).

5. Facing East, B's left turn changes her facing to North; walking 9 m North takes her from (11, 0) to (11, 9) — B's final position.

6. Subtract A's coordinates from B's coordinates: (11, 9) − (5, 9) = (6, 0), a purely horizontal difference.

Cross-check: both girls cover the same 9 m of northward travel overall (A's first leg, B's second leg), so the North-South coordinate is identical for both and cancels out — only the East-West coordinates need comparing. A stands 5 m East of the start and B stands 11 m East of the start, an 11 − 5 = 6 m gap in B's favour, matching the vector-subtraction result above.

B is therefore 6 m to the East of A.