Points: (2, 1) and (14, 6) Distance: 13
If you mean points of (-4, 2) and (1, 2) then the distance works out as 5
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
If you mean points of (-3, 1) and (-7, 1) then using the distance formula it is 10 units
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
11 points
Answer: 1
Points: (-6, 1) and (-2, -2) Distance: 5 units
(3-1)2 + (5-8)2 = 13 and the square root of this is the distance between the points
The distance between the points is two times the square root of 3.
Distance = (9-5)2+(-6-1)2 = 65 and the square root of this is the distance between the points which is about 8.062257748
Points: (2, 1) and (14, 6) Distance: 13
The shortest distance between the two points is zero
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
If you mean points of (2, 4) and (-1, 8) then the distance works as 5
If you mean points of (-4, 2) and (1, 2) then the distance works out as 5
If you mean points of: (2, 1) and (14, 6) then the distance is 13