I'm assuming that these are 3-D coordinates A(0,0,0) and B(3,4,5) The distance formula between two points in 3 dimensions is the following: D = SQR {(x1-x2)2 + (y1-y2)2 + (z1-z2)2} D = SQR {(0-3)2 + (0-4)2 + (0-5)2} D = SQR {9+16+25} = SQR {50} D is approx. = 7.07
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
Points: (0, 0) and (2, 6) Slope: 3
The sq.root of 122+162=20
If d is the distance between them, then d2 = (-6 -10)2 + (1 - (-8))2 = (-16)2 + 92 =256 + 81 = 337 so d = sqrt(337) = 18.36
Points: (2, 4) and (5, 0) Distance: 5
0 0
7
-2 & 5
Answer: 1
0 is the answer. If you start at -2 2, and end at -2 2, you moved 0 spots so there is no distance.
The distance between the points of (2, 3) and (7, 0) is the square root of 34
Distance between (0,0) and (8,2): √((8-0)2+(2-0)2) = √(82+22) = √(64+4) = √(68) = 2√17 ~= 8.246
It works out as the square root of 8 which is about 2.828 rounded to 3 decimal places
The square root of (2x2)+(5x5) is 5.385164807134504 ' . . . . . . .
Using Pythagoras: 5 squared + 2 squared = √21 which is about 4.58.
The distance between the points of (4, 3) and (0, 3) is 4 units