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If the square in the figure BELOW is rotated clockwise about th 06 Jun 2017, 03:40
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

45% (medium)

Question Stats:

65% (01:32) correct 35% (01:17) wrong based on 116 sessions

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If the square in the figure BELOW is rotated clockwise about the origin until vertex V is on the negative y-axis, then the new y-coordinate of V is

(A) –2
(B) −2 \(\sqrt{2}\)
(C) –4
(D) −3 \(\sqrt{2}\)
(E) –8

Source: Nova GMAT
Difficulty Level: 600
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B

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If the square in the figure BELOW is rotated clockwise about th 06 Jun 2017, 04:14
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SajjadAhmad
If the square in the figure BELOW is rotated clockwise about the origin until vertex V is on the negative y-axis, then the new y-coordinate of V is

(A) –2
(B) −2 \(\sqrt{2}\)
(C) –4
(D) −3 \(\sqrt{2}\)
(E) –8

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Distance of V from origin = \(\sqrt{{2^2+2^2}\) = \(2\sqrt{2}\)

The distance of V from origin will remain same as the Vertex V will follow the locus of a circle with radius \(2\sqrt{2}\)

Hence the co ordinate of V on negative Y axis = (0, \(-2\sqrt{2}\))

Answer: Option B
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If the square in the figure BELOW is rotated clockwise about th 06 Jun 2017, 04:32
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IMO B

Distance of v from origin is 2*sqrt(2)
So if it is rotated x coordinate will become 0 and y will be -2*sqrt(2)
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If the square in the figure BELOW is rotated clockwise about th 06 Jun 2017, 04:37
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B.The new vertex will be at the same distance from the origin and the old vertex. Which is sqrt 8 units. And only B satisfies it.

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If the square in the figure BELOW is rotated clockwise about th 06 Jun 2017, 06:20
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SajjadAhmad
If the square in the figure BELOW is rotated clockwise about the origin until vertex V is on the negative y-axis, then the new y-coordinate of V is

(A) –2
(B) −2 \(\sqrt{2}\)
(C) –4
(D) −3 \(\sqrt{2}\)
(E) –8

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In rotation problems, often it's easier to "see" what to track as a figure moves by drawing it. Connect V to Origin.

That's the line that whose rotated end point is the target. It is the distance from V to origin.

It is also the hypotenuse of a smaller isosceles right triangle in quadrant IV with side ratio of \(x : x : \sqrt{2}\). From (x2 - x1)*, side is 2. Hence hypotenuse is 2\(\sqrt{2}\), which = the absolute value of y on the rotated vertex, which will lie on the negative y-axis.

So y-coordinate is −2 \(\sqrt{2}\).

Answer B

* x2 is from V, x1 is from lower left vertex (0, -2) of small square in Q4
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If the square in the figure BELOW is rotated clockwise about th 16 Jun 2019, 23:02
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when we rotate the square to the negative y axis - it will become the diagonal of the square.

therefore we can answer this question within 10 seconds.

diagonal of square = sqrt2 * side of square.
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