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Solution

Facing north-east, your first move of 10 m forward takes you along a line at 45° from north. Then turning left from north-east → you face north-west direction, and you move 7.5 m. If you resolve the path: The first displacement = $(10 \text{m}, 45°)$ ⇒ Components: $x_1 = 10\cos45° = 7.07$, $y_1 = 10\sin45° = 7.07$ The second displacement = $(7.5 \text{m}, 135°)$ ⇒ Components: $x_2 = 7.5\cos135° = -5.30$, $y_2 = 7.5\sin135° = 5.30$ Net displacement components: $x = 7.07 - 5.30 = 1.77$, $y = 7.07 + 5.30 = 12.37$ Both $x$ and $y$ are positive ⇒ final position is north-east of initial position. But since the question asks relative to initial position along main axes, the major shift is northward.

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Solution

Coordinates: $(0,0)!\to!(0,-5)!\to!(-3,-5)!\to!(-3,-10)$, which is to the south-west of the start.

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Solution

Solution: Start at (0,0). After moves: (90,0) → (90,−20) → (60,−20) → (60,80). Distance = √(60²+80²) = √10000 = 100 m.