Distance formula: square root of (x1-x2)2+(y1-y2)2
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
(0, 4) and (- 4, 6) ???Distance = sqrt[(Y2 - Y1)2 + (X2 - X1 )2]Distance = sqrt[(6 - 4)2 + (- 4 - 0)2]Distance = sqrt( 4 + 16)Distance = sqrt(20)==============
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Points: (-1, -9) and (4, -2) Distance: (-1-4)2+(-9--2)2 = 74 and the square root of this is the distance which is about 8.602 to 3 decimal places
Points: (-1, -9) and (4, -2) Distance: (-1-4)2+(-9--2)2 = 74 and the square root of this is the distance which is about 8.602 to 3 decimal places
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 you mean points of (-4, 2) and (1, 2) then the distance works out as 5
If you mean points of (2, 4) and (-1, 8) then the distance works as 5
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
The distance between the points can be calculated by using the difference in the x coordinates, the difference in the y coordinates and Pythagoras. distance = sqrt((difference_in_x_coords)2 + difference_in_y_coords)2) So for the points (-1, 1) and (1, -1) the distance between them is: sqrt((-1 - 1)2 + (1 - -1)2) =sqrt(22 + 22) =sqrt(4 + 4) = sqrt(8) ~= 2.83
Points: (4, 4) and (-2, -2) Distance: 6 times square root of 2
Points: (2, 4) and (5, 0) Distance: 5
End points: (10, -4) and (2, 2) Midpoint: (6, -1) Distance from (6, -1) to (10, -4) = 5 Distance from (6, -1) to (2, 2) = 5 Equation of the circle: (x-6)^2 +(y+1)^2 = 25
square root of (5-9)^2+(1+6)^2
Using Pythagoras: distance = √(difference_in_x^2 + difference_in_y^2) = √((6 - 2)^2 + (3 - 4)^2) = √(16 + 1) = √17 ≈ 4.12