Distance formula from point to straight line:
|ax0+by0+c|/√(a^2+b^2)
For this problem, there are
|25X- 15Y+ 12|/5√34
Find the minimum value
Is the minimum value of |25X- 15Y+ 12|.
∵X and y are integer ∴25X- 15Y+ 12 divided by 5 and 2.
And 5-2 = 3 >; 2 (in order to avoid the phenomenon of 4, you need to write)
The minimum value of ∴|25X- 15Y+ 12| is 2.
It can be obtained when x=- 1 and y=- 1.
The minimum distance is 2/5√34=√(34)/8.