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

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

by Gmat_mission » Wed Apr 17, 2019 1:20 am

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A

B

C

D

E

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

Its center lies at (0, 0). If it is rotated clockwise by 45° around the center point, what will be the (x, y) coordinates of point D?

A. \((−2,0)\)
B. \((2,2)\)
C. \((−2,2)\)
D. \((−\sqrt{2},−\sqrt{2})\)
E. \((−\sqrt{2},\sqrt{2})\)

[spoiler]OA=E[/spoiler]

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by Vincen » Wed Apr 17, 2019 3:23 am
Hi Gmat_mission.

We can find the correct answer discarding the wrong answers.

First, because of the position of D after the rotation, we can conclude that before the rotation it was on the quadrant II. Therefore, the x-coordinate was negative and the y-coordinate was positive.

This allows us to discard the options A, B and D.

On the other hand, from the picture we can see that the distance from D to the origin must be 2 (it doesn't change after the rotation). Now, let's find the distance between C and O: $$d\left(C,O\right)=\sqrt{\left(-2-0\right)^2+\left(2-0\right)^2}=\sqrt{8}\ne2.$$ So, we discard C.

Hence, the correct answer is _E_. In additon,
$$d\left(E,O\right)=\sqrt{\left(-\sqrt{2}-0\right)^2+\left(\sqrt{2}-0\right)^2}=\sqrt{4}=2.$$

I hope it helps you.

Regards.
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by swerve » Wed Apr 17, 2019 9:22 am
\(D (x,y)\) has to satisfy: \(x^2+y^2=4\)

After rotate \(\Rightarrow\, y= -x\)

\(D\) is in II quadrant \(\Rightarrow\, x<0, y>0 \,\Rightarrow\,\) __E__ is the correct option.
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