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)==============
if ABCD is the triangle and O is the point then AO^2 + CO^2=BO^2+DO^2. Hence distance from 4th vertex can be calculated
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
2+2
Points: (2, 4) and (5, 0) Distance: 5
the distance 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)==============
Distance = sqrt[(Y2 - Y1)2 + (X2 - X1)2]Distance = sqrt[(9 - - 3)2 + (0 - 5)2]Distance = sqrt[(12)2 + (- 5)2]Distance = sqrt(144 + 25)Distance = sqrt(169)Distance = 13 units================
(Distance)2 = (2 - 5)2 + (6 - 2)2
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
What is the distance between (4, -2) and (-1,6)?
What is the distance between (4, -2) and (-1,6)?
0 is the answer. If you start at -2 2, and end at -2 2, you moved 0 spots so there is no distance.
Points: (-6, 1) and (-2, -2) Distance: 5 units
Time = Distance/Speed = 100/2 = 50 seconds.Time = Distance/Speed = 100/2 = 50 seconds.Time = Distance/Speed = 100/2 = 50 seconds.Time = Distance/Speed = 100/2 = 50 seconds.
Points: (4, 4) and (-2, -2) Distance: 6 times square root of 2
(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)==============