Find the distance from (6,2) to the line 2x - 3y = 6

I used this formula:
(Ax + By + C) / (sqrt. A^2 + B^2)

and got 0 for the numerator. Did I do the problem correctly?

Question

Find the distance from (6,2) to the line 2x - 3y = 6

I used this formula: 
(Ax + By + C) / (sqrt. A^2 + B^2) 

and got 0 for the numerator. Did I do the problem correctly?

raden · Answer

you just forgot the absolut sign, it is 
|(Ax + By + C) / (sqrt. A^2 + B^2)|

raden · Answer

first,
2x - 3y = 6 or 2x - 3y - 6 = 0
here A = 2, B = -3, and C = -6

raden · Answer

wait, it like be zero as you said, haha ...
|(2*6 - 3*2 - 6) / (sqrt. 2^2 + (-3)^2)| 
= |0/...|
= 0

raden · Answer

it means the point of (6,2) is on the line 2x - 3y = 6 , so no distance of them
|dw:1395372192407:dw|

raden · Answer

please recheck is the point given (6,2) or not ?

anonymous · Answer

Oh I believe it was supposed to be a trick question because the follow up question asks to explain the answer to this question. I didn't realize that (6,2) would be on the line. Thank you very much for your help!

raden · Answer

you're welcome :)