The distance between the points of (4, 3) and (0, 3) is 4 units
(3-1)2 + (5-8)2 = 13 and the square root of this is the distance between the points
I'm assuming that these are 3-D coordinates A(0,0,0) and B(3,4,5) The distance formula between two points in 3 dimensions is the following: D = SQR {(x1-x2)2 + (y1-y2)2 + (z1-z2)2} D = SQR {(0-3)2 + (0-4)2 + (0-5)2} D = SQR {9+16+25} = SQR {50} D is approx. = 7.07
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.
If you mean endpoints (-1, -3) and (11, -8) then by using the distance formula the length between the points is 13 units
Distance between the points of (3, 7) and (15, 16) is 15 units
The distance between the points of (4, 3) and (0, 3) is 4 units
Points: (3, -4) and (3, 3) Distance: 7 units
The distance between the points is two times the square root of 3.
Points: (8, 3) and (8, 6) Distance works out as: 3
√((7-3)² + (5 - -2)²) = √(4² + 7²) = √(16+49) = √(65) ≈ 8.062 ■
(3-1)2 + (5-8)2 = 13 and the square root of this is the distance between the points
The distance between points: (9, 4) and (3, 4) is 6
Points: (-5, -2) and (3, 13)Distance works out as 17 units
7.2111 (rounded)
The distance between the points of (2, 3) and (7, 0) is the square root of 34
If the points are (3, 2) and (9, 10) then the distance works out as 10