The sq.root of 122+162=20
If d is the distance between them, then d2 = (-6 -10)2 + (1 - (-8))2 = (-16)2 + 92 =256 + 81 = 337 so d = sqrt(337) = 18.36
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
Use Pythagoras' Theorem: calculate the square root of ((difference of x-coordinates)2 + (difference of y-coordinates)2).
If the points are (3, 2) and (9, 10) then the distance works out as 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
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.
If you mean points of: (-6, -10) and (2, 5) then it works out as 17
10 units
The sq.root of 122+162=20
The distance between (4, 5) and (10, 3) = sqrt(40) = 2*sqrt(10) = 6.3246 approx.
10
Points: (22, 27) and (2, -10) Distance using the distance formula: 42.06 rounded to two decimal places
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