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
If the points are (3, 2) and (9, 10) then the distance works out as 10
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-(-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