T starts from Point L and drives 3 km towards south. He then takes a left…

Direction-sense (distance-and-direction) problems are solved by placing every straight leg of the journey on a compass grid, East–West along one axis and North–South along the other, and tracking the walker's facing direction using the rule that a right turn rotates the facing 90° clockwise (North → East → South → West → North) while a left turn rotates it 90° counter-clockwise. Adding every leg as a vector on that grid gives the net displacement from the starting point to the stopping point; the shortest way back to the start is simply the reverse of that net displacement vector.

1. Let the starting point L be the origin (0, 0). Facing south, T drives 3 km, reaching (0, -3).

2. A left turn while facing south changes the facing to east. Driving 5 km east moves T from (0, -3) to (5, -3).

3. A right turn while facing east changes the facing to south. Driving 3 km south moves T from (5, -3) to (5, -6).

4. A right turn while facing south changes the facing to west. Driving 9 km west moves T from (5, -6) to (-4, -6).

5. A right turn while facing west changes the facing to north. Driving 8 km north moves T from (-4, -6) to (-4, 2).

6. A final right turn while facing north changes the facing to east. Driving 4 km east brings T to the stopping point M at (0, 2).

The net displacement from L to M is therefore (0, 2), that is, 2 km due north of L. To retrace the shortest path from M back to L, T must travel in exactly the reverse direction of that displacement: 2 km due south.

This can be checked independently by summing the east–west and north–south legs separately. East–west legs: +5 km (east) − 9 km (west) + 4 km (east) = 0 km, so there is no net east–west offset between L and M. North–south legs: −3 km (south) − 3 km (south) + 8 km (north) = +2 km, so M lies 2 km north of L. Both components agree with the coordinate result above, confirming that Point L lies 2 km due south of Point M.