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
(-3-5)2+(-1--1)2 = 64 and the square root of this is the distance which is 8
Points: (-4, 3) and (3, -1) Distance: (3--4)2+(-1-3)2 = 65 and the square root if this is the distance which is just over 8
Distance between (1, 3) and (-1, 0.5) = sqrt[(1 - -1)2 + (3 - 0.5)2] = sqrt(22 + 2.52) = sqrt(10.25) = 3.20 approx.
Points: (1, -2) and (1, -5) Distance: 3 units by using the distance formula
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
(-3-5)2+(-1--1)2 = 64 and the square root of this is the distance which is 8
3 and 1/2 miles
3 and 1/2 miles
(3-1)2 + (5-8)2 = 13 and the square root of this is the distance between the points
Points: (-4, 3) and (3, -1) Distance: (3--4)2+(-1-3)2 = 65 and the square root if this is the distance which is just over 8
4
The distance between (3, 7) and (-3, -1) is sqrt{[(3 - (-3)]2 + [7 - (-1)]2} = sqrt{[3 + 3]2 + [7 + 1]2} =sqrt{[6]2 + [8]2} = sqrt{36 + 64} = sqrt{100} = 10 units.
Distance between (1, 3) and (-1, 0.5) = sqrt[(1 - -1)2 + (3 - 0.5)2] = sqrt(22 + 2.52) = sqrt(10.25) = 3.20 approx.
Points: (1, -2) and (1, -5) Distance: 3 units by using the distance formula
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