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Use Pythagoras' Theorem: calculate the square root of ((difference of x-coordinates)2 + (difference of y-coordinates)2).

Q: Find the distance between the points -10 8 and -2 -7?

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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

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

Related questions

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.

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.

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

The distance between (4, 5) and (10, 3) = sqrt(40) = 2*sqrt(10) = 6.3246 approx.