This means that point A lies on the perpendicular bisector of the line joining points O and B.
Let's find that perpendicular bisector.

The line passing through O and B is y = x. This is very much visible by point (0,0) and (4,4), saving us some calculation here.
Hence, the perpendicular line must be y = -x + C, in which the slope = -1, negative reciprocal of slope = 1 of the line y = x. The y-intercept of this line is C. We can find out C by plugging in coordinates of a point passing through this line.

The perpendicular bisector must be passing through (2,2). You can simply find this by applying the distance formula or by averaging the coordinates of O and B.
Now let's use this point (2,2) to find C in y = -x + C.

2 = - 2 + C
C = 4
Hence, the perpendicular bisector is y = -x + 4.

You may find plugging in the answer choices in this equation to find the possible value of A. But there will be more than one such answers.


Let's use this information and the distance formula to find A on the perpendicular bisector we have found.

Area\( = 16 = \frac{1}{2} * 4\sqrt{2} * D\), in which D is the distance of point A from OB.
\(D = 4\sqrt{2}\)

Now, applying distance formula, we can find the coordinates, say, (r,s) of point A.
\(\sqrt{(r-2)^2 + (s-2)^2} = 4\sqrt{2}\)
\((r-2)^2 + (s-2)^2 = 16 * 2\)

We have 2 variables, but 1 equation. Really? No. We also have the perpendicular bisector on which these coordinates lie.
s = -r + 4

Putting the values of s in equation \((r-2)^2 + (s-2)^2 = 16 * 2\), let's solve the equations simultaneously.
\((r-2)^2 + (-r + 4 -2)^2 = 32\)
\((r-2)^2 + (-r + 2)^2 = 32\), but inside the square (2-r) = (r-2)
\(2 * (r -2)^2 = 32\)
\(r-2 = ± 4\)
r = 6, -2
s = -2, 6
So, point A can be either (6, -2) or (-2, 6).
Hence, A is the correct answer.