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If yoiu mean pt A = ( 8,2) & B = (3,8)
We apply Pythagoras.
Distance 'd^(2)' = (x(A) - x(B))^2 + (y)A) - y(B))^(2)
Substituting
d^)2 = (8 - 3)^(2) + ( 2 - 8)^(2)
d^(2) = 5^(2) + -6^(2)
d^(2) = 25 + 36
d^(2) = 61
d = sqrt(61)
Square root of 61.
Distance formula is: square root of ((x1-x2)squared + (y2-y1)squared))
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Just subtract the lowest number from the greatest number. For example, the distance between 3 and 8, is 8 - 3 = 5 units, the distance between -2 and 3, is 3 - (-2) = 3 + 2 = 5 units, the distance between -4 and -2, is -2 - (-4) = -2 + 4 = 2 units.
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What is the distance between points 1 2 and 4 -1?
Use Pythagoras to find the distance between two points (x0,.y0) and (x1, y1): distance = √(change_in_x² + change_in_y²) → distance = √((x1 - x0)² + (y1 - y0)²) → distance = √((4 - 1)² + (-1 -2)²) → distance = √(3² + (-2)²) → distance = √(9 + 9) → distance = √18 = 3 √2
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If you mean points of (1, -2) and (-9, 3) then the distance is about 11 units using the distance formula
