The Pythagorean distance is sqrt[(4 - 7)2 + (7 - 8)2] = sqrt[9 + 1] = sqrt(10) = 3.162 approx.
(-3-(-6))2 + (7-4)2 = 18 and the square root of this is the distance between the two points
The distance between the points of (4, 3) and (0, 3) is 4 units
Using the distance formula from (3, 1) to (7, 1) is 4 units
The distance works out as 22 between the points of (15, -17) and (-7, -17)
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.
(-3-(-6))2 + (7-4)2 = 18 and the square root of this is the distance between the two points
Points: (3, -4) and (3, 3) Distance: 7 units
45
The distance between the points of (4, 3) and (0, 3) is 4 units
Points: (2, 3) and (2, 7) Distance works out as: 4 units
If you mean points of (-2, 4) and (5, 4) then using the distance formula it is 7
Using the distance formula from (3, 1) to (7, 1) is 4 units
To find the distance between the points (7, 5) and (4, 9), you can use the distance formula: ( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ). Plugging in the values, ( d = \sqrt{(4 - 7)^2 + (9 - 5)^2} = \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 ). Therefore, the distance between the two points is 5 units.
If you mean points of (-5, -3) and (7, 7) then by using the distance formula it is 120 units
√((7-3)² + (5 - -2)²) = √(4² + 7²) = √(16+49) = √(65) ≈ 8.062 ■
Distance between the points of (3, 7) and (15, 16) is 15 units
The distance between the points is two times the square root of 3.