Distance calculations

[not bright]

I am trying to get a formula to produce a distance in feet
Example:
A B C
X corrd y corrd
1 6806.9 1680.62
2 7243.67 1509.92 (Dist. Between a1 & a2)
3 6751.07 1319.53 (Dist. Between a2 & a3)

The numbers are X & Y Coordinates for devices in a CAD program.

I have a formula that I was trying: (ABS(A1-A2)+ABS(B3-B2))/12)
But the results don’t match to the distance in the CAD drawing. Or, are my
math skills that bad???

 

[Otto Moehrbach]

Assuming Column A is the "X" coordinates of points and Column B is the "Y"
coordinates of the same points, as you have them, then the straight line
distance
between the two points, rows 1 and 2, is the hypotenuse of a right triangle
as follows:
One leg is Abs(A2-A1)
The other leg is Abs(B2-B1)
The hypotenuse is the square root of the sum of the squares of the other two
sides.
The hypotenuse is (((Abs(A2-A1))^2 + (Abs(B2-B1))^2)^.5)/12
HTH Otto

 

[Gary''s Student]

I am sure that you are very bright, maybe a little forgetful:

Remember Pythagoras

D=sqrt((x1-x2)^2+(y1-y2)^2))

 

[not bright]

Thanks alot

does it matter which one is first (x1-x2 or x2-x1)

 

[Bernard Liengme]

Since you are squaring the order is unimportant.
3-5 = -2 and -2^2 =4
5-3 = 2 and 2^2 =
best wishes