Points: (2, 3) and (2, 7) Distance works out as: 4 units
Points: (-4, 3) and (3, -1) Distance: (3--4)2+(-1-3)2 = 65 and the square root if this is the distance which is just over 8
The distance between the points of (2, 3) and (7, 0) is the square root of 34
Use Pythagoras to find the distance between two points (x0,.y0) and (x1, y1): distance = √(change_in_x² + change_in_y²) → distance = √((x1 - x0)² + (y1 - y0)²) → distance = √((4 - 1)² + (-1 -2)²) → distance = √(3² + (-2)²) → distance = √(9 + 9) → distance = √18 = 3 √2
Points: (1, -2) and (1, -5) Distance: 3 units by using the distance formula
(3-1)2 + (5-8)2 = 13 and the square root of this is the distance between the points
Points: (-5, -2) and (3, 13)Distance works out as 17 units
Points: (2, 3) and (2, 7) Distance works out as: 4 units
(-3-(-6))2 + (7-4)2 = 18 and the square root of this is the distance between the two points
Points: (-4, 3) and (3, -1) Distance: (3--4)2+(-1-3)2 = 65 and the square root if this is the distance which is just over 8
If the points are (3, 2) and (9, 10) then the distance works out as 10
The distance between the points of (2, 3) and (7, 0) is the square root of 34
Use Pythagoras to find the distance between two points (x0,.y0) and (x1, y1): distance = √(change_in_x² + change_in_y²) → distance = √((x1 - x0)² + (y1 - y0)²) → distance = √((4 - 1)² + (-1 -2)²) → distance = √(3² + (-2)²) → distance = √(9 + 9) → distance = √18 = 3 √2
Points: (1, -2) and (1, -5) Distance: 3 units by using the distance formula
If you mean points of: (-5, 1) and (-2, 3) then the distance is about 3.61 rounded to two decimal places
The distance between the points of (4, 3) and (0, 3) is 4 units
If you mean points of (1, -2) and (-9, 3) then the distance is about 11 units using the distance formula