A square is drawn on the xy coordinate plane as shown:

As the point D would lie in the 2nd quadrant, point (x,y) would have a (-,+) sign; which gets rid of option B and D.
As D is moving by 45 degrees, it's Y coordinate cant be 0 which eliminates point A.

Distance from center to point A can be calculated as 2 units which would be constant for all 4 points and only point E satisfies that.

They key to solving this question is understanding what constitutes a 45 degree clockwise rotation and how that actually changes the position of point "D." More fundamentally, if we rotate the figure 45 degrees clockwise then the X coordinate for point D would be equal to negative half of the length of line AD. The trap in this question is essentially getting test takers to think that because point "O" is the origin, then the length of AD, BA, BC, DC as the figure resembles a rotated square. However, this is not actually the case. If we know the length of triangle AOD's hypotenuse then we calculate x and y coordinate of point. Moreover, this question also presents answer chances that can be quickly "eyeballed" and eliminated provided the test taker knows the properties of each quadrant of the graph ( example all x and y values in quadrant iii must be negative). Anyways, point O" is equidistant from point A and point D- therefore, the length of both AO and AD is 2. If we use the Pythagoren Theorem then we can solve for the length of side AD, the hypotenuse.

Length of Side AD (hypotenuse)

2^2 + 2^2 = 8
\sqrt{8}=
(4)\sqrt{2}=
2\sqrt{2}

Now if we divide this length by two and convert it to negative, then we have the x value of "D"

-\sqrt{2}/2


According to the properties of Quadrant III (which we already know D must be in by definition of a 45 degree clockwise rotation), answer choices D, B and A clearly cannot be the coordinates of D. The only two choices are "C" and "E." Knowing that point D's x value must be -\sqrt{2} the answer must be E.

Hence "E"

D (x,y) has to satisfy: x^2+y^2=4
After rotate => y= -x
D is in II quadrant => x<0, y>0 => E is the answer

Hi
Can you please elaborate the solution a bit more?

Hi All,

This question can actually be solved with no math at all, but you will need to understand the logic behind the prompt.

Since the square is centered around the Origin, Point D is at (-2,0). IF we rotated the square 90 degrees clockwise, it would be where Point is now (0,2). But the question asks us to rotate it just 45 degrees clockwise, so Point D will land in the 2nd Quadrant (with a NEGATIVE X -coordinate and a POSITIVE Y-coordinate). Eliminate Answers A, B and D.

In addition, rotating the square a full 360 degrees will send Point D through a complete CIRCLE, so you could do a rough sketch of what that would look like. You'll see that answer C would fall outside of that circle, so it can't be the answer. Eliminate C.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Thanks Sir for a wonderful approach to the question.
I think there is a typo in the solution, D is will be in second quadrant not tge third.

Please confirm my understanding.

Question that reminds you to always glance at the answer choices for clues before you start any question.

1st) Rule: the Diagonals of a Square are Perpendicular Bisectors of Each other. The 2 Diagonals will cross at the Center of the Square and create a Perpendicular Angle and Bisect Each Other

Since the Center of the Square is at Origin (0 , 0) and 90 Degree Angles are created at the Center by Lines AC and DB ---- these must the Diagonals Bisecting the Vertices and themselves in the Square

the 2 Diagonals go from Point D to Point B --- and --- Point C to Point A


Again, because the Diagonals are Bisectors, along the Diagonal Lines:

AO = OC
and
DO = OB


This means Each Vertex must undoubtedly be 2 Units away from the Origin, Since Point A is 1/2 a Diagonal and 2 Units away from the Origin/Center

Point D is at (-2 , 0)
Point B is at (2 , 0)
Point C is at (0 , -2)


2nd) Each Exterior Angle of a Square = 360 deg./ 4 Equal Exterior Angles = 90 Degrees

Thus, a 90 Degree Clockwise Movement would result in Point D switching places with Point A at (0, 2)

a 45 Degree Clockwise Movement is 1/2 that Movement. Point D will therefore move on to the Line given by the Equation of: y = (-)X.

Point D will keep the Same Distance from the Origin = 2. Also, the Line Y = (-)X creates 45 Degree Angles with the X-Axis and Y-Axis.

At this Point, you can notice that TWO 45/45/90 Degree Triangles can be formed with Point D as the Vertex and the Length of DO = 2 as the Hypotenuse. You can find the exact coordinates of D that way


Or you can observe that Point D will be in Quadrant II, with a (-)Negative X Value and a (+)Y Value

this Eliminates every answer except C and E


Lastly, without doing any calculations, you should be able to Observe that Answer C's Coordinates of (-2 , 2) would incorrectly INCREASE the Length of OD = 2 on the Graph Paper.

The Answer must be E

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