Habi travelled from a point X straight to Y at a distance of 80 metres. He…

Concept: In a turning-path (direction-sense) problem, only movements that fall on the SAME straight-line axis add up or cancel algebraically; every 90° turn switches the axis of travel. Two segments that end up parallel but pointing in opposite directions cancel each other completely. The net displacement from the start is found by plotting the (x, y) position after every turn and then measuring the straight-line distance between the start and the final point.

Applying it here: Fix X as the origin and let X→Y run along one axis (say, north); each “turn right” rotates the direction of travel 90° clockwise:

1. X to Y: walk 80 m north, reaching Y = (0, 80).

2. At Y, turn right (north → east) and walk 50 m, reaching A = (50, 80).

3. At A, turn right (east → south) and walk 70 m, reaching B = (50, 10).

4. At B, turn right (south → west) and walk 50 m, reaching C = (0, 10).

Cross-check: The two 50 m east–west segments (Y→A and B→C) are equal and opposite, so they cancel exactly on that axis; the north–south axis nets to 80 − 70 = 10 m. This gives the straight-line distance XC = 10 m, and the result holds regardless of which compass direction X→Y is assumed to point in — only the relative turns matter.

So Habi ends up 10 metres from the starting point X.