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

Question

https://gmatclub.com/forum/download/file.php?id=53361 In the rectangular coordinate system above, if point A (not shown) is equidistant from points O and B and the area of triangle OAB is 16, which of the following are the possible coordinates of point A? A. (-2, 6) B. (0, 4) C. (2, -6) D. (2, 6) E. (4, 0) 1 (1).png

Bunuel · Accepted Answer

Look at the diagram below: https://gmatclub.com/forum/download/file.php?id=20291 Since A is equidistant from points O and B, then it must be somewhere on the green line (perpendicular bisector of OB). (2,-6) and (2,6) are not on that line. If A is at (0,4) or (4,0), then the area is 1/2*4*4=8. Thus A must be at (-2,6). Answer: A. Hope it's clear. Area.png

aa008 · Answer

Area of a triangle in a Co-ordinate system= \frac{1}{2}  
let the third co-ordinate be (x,y)

so  (x1,y1)=(x,y) 
  (x2,y2)=(4,4) 
  (x3,y3)=(0,0)

on putting these value in the formula above, we get

area= \frac{1}{2}  
 = 2(x-y)

area is already given so,
 
 2(x-y)=16 
 x-y=8

it mean that distance between 2 point on number line is 8
with this knowledge in mind if we scan the answer choice we are left with only two choices: Choice A(-2,6) or Choice C(2,-6)

Now, if you apply your logical sense, choice C is not equidistant from both point, but choice A is.

A is the correct answer.

mau5 · Answer

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) . Only A satisfies for h =  4\sqrt{2} 
A.

AmritaSarkar89 · Answer

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

GMATinsight · Answer

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

chetan2u · Answer

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

Fdambro294 · Answer

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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rahulbanka · Answer

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

Foreheadson · 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.

bumpbot · Answer

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

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

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