A man leaves from his home. He first walks 40 meters in north-west direction and then 40 meters in south-west direction. Next he walks 40 meters in south-east direction. Finally, he turns towards his house. In which direction is he moving?

Let the starting position (home) be at the origin (0, 0).

Taking east as positive x-direction and north as positive y-direction.

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After walking 40 meters in north-west direction:

North-west makes 45° with both north and west axes.

x-displacement: $-40 \cos(45°) = -40 \times \dfrac{1}{\sqrt{2}} = -20\sqrt{2}$

y-displacement: $40 \sin(45°) = 40 \times \dfrac{1}{\sqrt{2}} = 20\sqrt{2}$

Position: $(-20\sqrt{2}, 20\sqrt{2})$

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After walking 40 meters in south-west direction:

South-west makes 45° with both south and west axes.

Additional x-displacement: $-40 \cos(45°) = -20\sqrt{2}$

Additional y-displacement: $-40 \sin(45°) = -20\sqrt{2}$

Position: $(-20\sqrt{2} - 20\sqrt{2}, 20\sqrt{2} - 20\sqrt{2}) = (-40\sqrt{2}, 0)$

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After walking 40 meters in south-east direction:

South-east makes 45° with both south and east axes.

Additional x-displacement: $40 \cos(45°) = 20\sqrt{2}$

Additional y-displacement: $-40 \sin(45°) = -20\sqrt{2}$

Final position: $(-40\sqrt{2} + 20\sqrt{2}, 0 - 20\sqrt{2}) = (-20\sqrt{2}, -20\sqrt{2})$

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To return home, he needs to go from $(-20\sqrt{2}, -20\sqrt{2})$ to $(0, 0)$.

Required displacement:

x-direction: $0 - (-20\sqrt{2}) = 20\sqrt{2}$ (positive, towards east)

y-direction: $0 - (-20\sqrt{2}) = 20\sqrt{2}$ (positive, towards north)

Since both displacements are equal and positive, the direction is north-east.

Therefore, the answer is Option 1: North-east.