The sq.root of 122+162=20
Use Pythagoras' Theorem: calculate the square root of ((difference of x-coordinates)2 + (difference of y-coordinates)2).
square root of ((22 -2)^2 + (27 +10)^2) = square root of (400 + 1369) which is equal to square root of (1769) = 42.05948169
A scale where the distance between 1 and 2 is the same as the distance between 2 and 3; and so on. By way of contrast, in a logarithmic scale, the distance between 1 and 10 would be the same as the distance between 10 and 100, or 100 and 1000. There are many other scales.
40
If the points are (3, 2) and (9, 10) then the distance works out as 10
10
The sq.root of 122+162=20
10
If you mean points of (4, 5) and (10, 13) then the distance works out as 10
Points: (2, 2) and (8, -6) Distance: 10
The distance between points can be calculated using Pythagoras: distance = √(change_in_x² + change_in_y²) → distance = √((4 - 10)² + (36 - 12)²) → distance = √((-6)² + (24)²) → distance = √(36 + 576) → distance = √612 → distance = 6 √17 ≈ 24.74 units.
The distance along a straight line is 10. Using the Pythagorean equation, c2 = a2 + b2 where the x change is 6 and the y change is 8, c2 = 62 + 82 = 36 + 64 = 100 c = [sqrt 100] = 10
If you mean points of: (-6, -10) and (2, 5) then it works out as 17
10 units
(-10--2)2+(8--7)2 = 289 and the square root of this is 17 Therefore the distance is 17 units in length.
The distance between (4, 5) and (10, 3) = sqrt(40) = 2*sqrt(10) = 6.3246 approx.