The answer will be the diagonal (hypotenuse) for a horizontal distance x2-x1 (4) and a vertical distance y2-y1 (4). The square root of the squares is sqrt [42 + 42] = sqrt [32] = approx 5.66
0 0
8.54
Points: (2, 4) and (5, 0) Distance: 5
Points: (-5, -2) and (3, 13)Distance works out as 17 units
The distance between two points is Square root of [ (difference in their 'x' coordinates)2 + (difference in their 'y' coordinates)2 ]
0 0
8.54
10
If the points are (3, 2) and (9, 10) then the distance works out as 10
the distance between two points is length
The distance between the points of (4, 3) and (0, 3) is 4 units
11 points
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
Points: (-6, 1) and (-2, -2) Distance: 5 units
Points: (2, 2) and (8, -6) Distance: 10
The 3-D distance formula depends upon what the two points are that you are trying to find the distance between. In order to find the formula, you need to enter 2 sets of coordinates in the 3 dimensional Cartesian coordinate system, and then calculate the distance between the points.
the distance between 2 points