Pythagoras can be used to find the distance between any two points (x0, y0) and (x1, y1):
Distance = √((x1 - x0)² + (y1 - y0)²)
→ distance = √((5 - 5)² + (13 - 9)²) = √(0 + 4²) = √(4²) = 4.
4
To find the distance between two points (x0, y0) and (x1, y1) use Pythagoras: distance = √(change_in_x² + change_in_y²) → distance = √((x1 - x0)² + (y1 - y0)²) → distance = √((8 - -7)² + (-5 - -13)²) → distance = √(15² + 8²) → distance = √289 → distance = 17 units.
If you mean: (-8, -3) and (4, 5) then the distance is 4 times square root 13
Distance = (9-5)2+(-6-1)2 = 65 and the square root of this is the distance between the points which is about 8.062257748
1
If you mean points of: (-5, 1) and (-2, 3) then the distance is about 3.61 rounded to two decimal places
Points: (-5, -2) and (3, 13)Distance works out as 17 units
If you mean: (-8, 4) and (5, 4) Then the distance between the points works out as 13
(3-1)2 + (5-8)2 = 13 and the square root of this is the distance between the points
Points: (-4, 5) and (3, 16) Distance: square root of 170 which is about 13
If you mean points of (4, 5) and (10, 13) then the distance works out as 10
To find the distance between two points (x0, y0) and (x1, y1) use Pythagoras: distance = √(change_in_x² + change_in_y²) → distance = √((x1 - x0)² + (y1 - y0)²) → distance = √((8 - -7)² + (-5 - -13)²) → distance = √(15² + 8²) → distance = √289 → distance = 17 units.
10 units
(2-5)2+(5-7)2 = 13 and the square root of this is the distance which is about 3.60555 to 5 decimal places
the distance between two points is length
Points: (2, 4) and (5, 0) Distance: 5
If you mean: (-8, -3) and (4, 5) then the distance is 4 times square root 13
If you mean points of (5, 5) and (1, 5) then the distance is 4