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.
Points: (2, 1) and (14, 6) Distance: 13
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
The shortest distance between the two points is zero
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
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, 1) and (14, 6) then the distance is 13