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Points: (6, -2) and (6, 2)Using the distance formula: 4
6 - the same as from 4 to 10.
One-half mile.
Using Pythagoras: distance = √(difference_in_x^2 + difference_in_y^2) = √((6 - 2)^2 + (3 - 4)^2) = √(16 + 1) = √17 ≈ 4.12
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