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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.

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Q: What is the distance between the points 5 9 and 5 13?

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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

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If you mean points of: (-5, 1) and (-2, 3) then the distance is about 3.61 rounded to two decimal places

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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