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Use Pythagoras:

distance = √(difference_in_x^2 + difference_in_y^2)

= √((6 - -3)^2 + (2 - -2)^2)

= √(9^2 + 4^2)

= √(81 + 16)

= √97 ≈ 9.85 units

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Q: What is the distance between 6 2 and -3 -2?
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