well, we'll be using the distance formula twice here.
that point, is on the x-axis, what's the value of "y" at the x-axis? well y = 0, since it's way down at the 0 level, so the point must be (x , 0).
the distance then from (x,0) to (-3,-1) is the same distance as from (x,0) to 1,6).
