Starting from a point 'M', Hari walked 18 metres towards south. He turned to…

Concept: In direction-sense problems, every walk is resolved onto two perpendicular axes: North-South and East-West. A left turn rotates the walker's facing by 90° in a fixed way (South → East → North → West → South, and so on), and the final displacement from the start is obtained by summing the movements along each axis separately — two equal-and-opposite movements on the same axis cancel completely, leaving only the net remainder to fix the true distance and direction.

Applying it to this walk:

1. Hari starts at M and walks 18 m South.

2. Facing South, a left turn points him East; he walks 25 m East.

3. Facing East, a left turn points him North; he walks 18 m North.

4. Facing North, a left turn points him West; he walks 35 m West and reaches P.

Resolving the two axes:

• North-South axis: 18 m South and 18 m North are equal and opposite, so they cancel completely — net vertical displacement is zero.

• East-West axis: 25 m East and 35 m West are opposite; the final westward leg is 10 m longer, so the net horizontal displacement is 10 m in its direction.

Cross-check: Taking M as the origin with East as +x and North as +y: after the four walks the position moves from (0, 0) to (0, −18) to (25, −18) to (25, 0) to (−10, 0) — confirming a single net displacement of 10 m along the horizontal axis, matching the diagram.

Hari (point P) is 10 m west of the starting point M.