8.06226 is. (rounded)
Points: (-5, -2) and (3, 13)Distance works out as 17 units
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
To find the distance between the points (3, -8) and (3, -19), you can use the distance formula, which is (d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}). Substituting the points into the formula: (d = \sqrt{(3 - 3)^2 + (-19 - (-8))^2} = \sqrt{0 + (-11)^2} = \sqrt{121} = 11). Thus, the distance between the points is 11.
(3-1)2 + (5-8)2 = 13 and the square root of this is the distance between the points
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
If the points are (3, 2) and (9, 10) then the distance works out as 10
Points: (1, -2) and (1, -5) Distance: 3 units by using the distance formula
(-3-(-6))2 + (7-4)2 = 18 and the square root of this is the distance between the two points
If you mean points of: (-5, 1) and (-2, 3) then the distance is about 3.61 rounded to two decimal places
If you mean points of (1, -2) and (-9, 3) then the distance is about 11 units using the distance formula
(-2-3)2+(7-7)2 = 25 and the square root of this is the distance which is 5