I'm not sure if this is the method that you want to solve this with, although it does use derivatives.

Use the distance formula to minimize the distance. One point is (0,0) and the other is going to be some point (x,y) on the line, so we can use that fact to replace y with 10-2x. So the two points are (0,0) and (x,10-2x).

The distance formula solves for D and has a square root, but you should notice that if you minimize D^2 = D*D, then you minimize D as well and it makes the equation much easier to work with.

So:

Now you differentiate and find the critical x-value, plug that into y=10-2x and you should have your point.