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.
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.
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
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
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.
If you mean points of (4, 5) and (10, 13) then the distance works out as 10
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 points of (5, 5) and (1, 5) then the distance is 4
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