In the rectangular coordinate system above, if point A (not shown) is

Assuming the base of the triangle as OB, which is of length \(4\sqrt{2}\), and let the height from A to OB be h-->

\(\frac{1}{2}*4\sqrt{2}*h\) = 16 --> h = \(4\sqrt{2}\). Also, as the point A is equidistant from both O and B, the point A will lie on the perpendicular bisector of the triangle OAB. Thus, h = the distance between the co-ordinates of A and (2,2)[the mid point of the line segment OB]. Only A satisfies for h = \(4\sqrt{2}\)
A.

Another way of solving this.

When we draw the given senerio we realise for A to be equidistant to both o and B it needs to be on the 2nd or 4th quadrant. only option in the second quadrant is A

hence A

There is an option which shows a point in 4th Quadrant (which one should notice even if one goes by quick observation like you have shown) however that is much closer to O than point B so that can't be the correct option.

Your approach is really good. exactly like one needed in any aptitude test. Overlooking option would be too bad for anyone taking GMAT

Look at the attached figure...
The point exactly in centre of O and B will be (2,2) and all points on the line that is perpendicular to OB will be at equal distance from both O and B.

First let us see which all line can be on it..
The slope of OB will be (4-0)/(4-0)=1, so it's perpendicular should have slope -1..

(A) (−2, 6)....slope (6-2)/(-2-2)=4/-4=-1.... possible
(B) (0, 4).....slope (4-2)/(0-2)=-1... possible
(C) (2, −6)...slope (-6-2)/(2-2)....NO
(D) (2, 6).....slope (6-2)/(2-2)...NO
(E) (4, 0)......slope (0-2)/(4-2)=-1... possible

So three options left, check for area ..
A) (-2,6)...
So triangle is (0,0);(-2,6);(4,4)
Area ={ √(4^2+4^2)*√((6-2)^2+(-2-2)^2}/2=(√32*√32)/2=32/2=16...yes
B) (0,4)
So sides are X=4 and y=4
Area = 4*4/2=8..no
Similarly for E

Ans A

Concept: the line that is equidistant from the origin and Point (4 , 4) will lie on the perpendicular Bisector of the line segment connecting those 2 points.

The midpoint between the origin and (4 , 4) is given by: (2 , 2)

The slope of line segment OB = +1

The perpendicular Bisector will have the slope that is the negative reciprocal of +1———(-)1

After plugging in the midpoint (2, 2) into the equation of: y = (-)1x + b

The formula for the perpendicular Bisector of segment OB, in which every point will be equidistant from the two points, is given by

Y = (-)1X + 4

You can then plug in the answer choices to see which points satisfy the above equation.

Only answers A and E satisfy the equation and lies on this perpendicular bisector line.

Without doing anymore working and estimating the area, you could choose A

However, you could also see that the Base OB = 4 * sqrt(2) —— because the line segment given by OB would be the diagonal of a 4 by 4 square


Area = (1/2) * 4 * sqrt(2) * H = 16

H = 4 * sqrt(2)

Thus, from mid-point (2 , 2) we are looking for a point that is the opposite diagonal vertex of a 4 by 4 square.

(6 , -2) or (-2, 6) would do the job.

A

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Any point lying on the perpendicular bisector of two points is equidistant from the two points.

Equation of perpendicular bisector of points O and B is Y = 4 – X
Only answer options A, B, and E satisfy this equation
Of these, options B and E don’t form a triangle with an area of 16
So, A is the answer

You have to realize that equidistant will be something on the line y=-x+4. Whichever answer choice satisfied this equation is the answer. in this case answer choice A is the correct one.

Bunuel


how did you find equidistant points (2,-6) and (2,6) just by looking at the green?