Quote:
point A (not shown) is equidistant from points O and B
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.
Quote:
the area of triangle OAB is 16
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.