|AB| = sqrt[(5 - 2)2 + (7 - 4)2] =sqrt[9 + 9] = 3*sqrt(2)
Since they are the same point, the distance between them is 0.
Distance = sqrt [(Y2 - Y1)2 + (X2 - X1)2]Distance = sqrt [(6 - 4)2 + (- 4 - 0)2]Distance = sqrt [(2)2 + (- 4)2]Distance = sqrt(4 + 16)Distance = sqrt(20)==============
Points: (2, 4) and (5, 0) Distance: 5
The distance is 2, along the x-value line x=3 i.e. (3, y).
There is no difference in the y-coordinates so the distance is simply in the x-coordinates and that is |-4 -4| = |-8| = 8
sqrt[(-4 - 4)2 + (-6 - 2)2] = sqrt[82 + 82] = sqrt(64 + 64] = sqrt(128) = 11.31 approx
What is the distance between (4, -2) and (-1,6)?
If you mean points of (5, 5) and (1, 5) then the distance is 4
What is the distance between (4, -2) and (-1,6)?
Since they are the same point, the distance between them is 0.
What is the distance between (4, -2) and (-1,6)?
The distance between the starting point and the destination is 143mi, (230km), and will take approximately 2 hours 4 minutes of driving time.
It is the square root of (-6-4)2+(1-3)2 = 2 times sq rt of 26 or about 10.198 to 3 decimal places
Using Pythagoras: distance = √(change_in_x2 + change_in_y2) = √((5 - -8)2 + (4 - 4)2) = √(132 + 02) = √(132) = 13 units.
Distance = sqrt [(Y2 - Y1)2 + (X2 - X1)2]Distance = sqrt [(6 - 4)2 + (- 4 - 0)2]Distance = sqrt [(2)2 + (- 4)2]Distance = sqrt(4 + 16)Distance = sqrt(20)==============
The distance between the origin (0, 0) and the point (4, -6) can be calculated using the distance formula: ( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ). Substituting the coordinates, we get ( d = \sqrt{(4 - 0)^2 + (-6 - 0)^2} = \sqrt{16 + 36} = \sqrt{52} ). Rounding to two decimal places, the distance is approximately 7.21.
The line ( y = 3 ) is a horizontal line. The distance from the point ( (5, 4) ) to this line can be found by calculating the vertical distance between the point and the line. Since the y-coordinate of the point is 4 and the line is at ( y = 3 ), the distance is ( |4 - 3| = 1 ). Therefore, the distance from the point ( (5, 4) ) to the line ( y = 3 ) is 1 unit.