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
Simply divide the difference in the y-coordinates, by the difference in the x-coordinates.
Passing through the points of -8-0 and 1-5 Difference between y coordinates of first and second points: =(-0) - (-5) = +5 Difference between x coordinates of first and second points: =(-8) - (1) = -9 So slope = 5/(-9) = -5/9
If you mean points of: (0, 3) and (8, -4) then it works out as -7/8
the numbers between 0 and 1 is 0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,0.10.
0 0
Since they are the same point, the distance between them is 0.
The distance between the points of (4, 3) and (0, 3) is 4 units
Points: (2, 4) and (5, 0) Distance: 5
To find the distance between the origin and the point (x,y) use Pythagoras on the right angled triangle which has the points (0, 0), (x, 0), (x, y) - the distance is the hypotenuse of the triangle and so has length: distance = √(x2 + y2) This can be extended to find the distance between any two points (x1, y1) and (x2, y2): distance = √((x2 - x1)2 + (y2 - y1)2) (for the original question (x1, y1) is the origin (0, 0) and the first formula results.)
5 is.
1
7
The distance between the points of (2, 3) and (7, 0) is the square root of 34
9
Answer: 1
Pythagoras can be used to find the distance between any two points (x0, y0) and (x1, y1): Distance = √((x1 - x0)² + (y1 - y0)²) → distance = √((5 - 5)² + (13 - 9)²) = √(0 + 4²) = √(4²) = 4.