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
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
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
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