Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.
1) Point D is 2 units away from origin 2) the square is rotated by 45 degree..
INFERENCE
1) point D will remain 2 units away from origin.. 2) since it is rotated by 45 degree, D will move on the bisector of both axis, that is D will be equidistant from two axis..
Solution
It becomes an Isosceles right angle triangle with Origin to D as the hyp.. so \(sides= hyp/\sqrt{2}=2/\sqrt{2}=\sqrt{2}\) Since D moves to III quad, x will be -ive and y positive.. ans -\(\sqrt{2},\sqrt{2}\) E
_________________
A square is drawn on the xy coordinate plane as shown:
[#permalink]
Show Tags
09 Mar 2016, 00:22
2
2
Another way of looking at the question:
We can tell that the diagonal = 4 Hence the side = 4/ \(\sqrt{2}\) = 2 \(\sqrt{2}\)
The X and Y axes will bisect the sides and the figure will be something like this after the rotation:
Attachment:
square rotated.JPG [ 12.54 KiB | Viewed 2118 times ]
Hence on each side of the Axes, there would be \(\sqrt{2}\) length. Hence the co ordinates of the point after rotation = (- \(\sqrt{2}\) , \(\sqrt{2}\) )
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.
1) Point D is 2 units away from origin 2) the square is rotated by 45 degree..
INFERENCE
1) point D will remain 2 units away from origin.. 2) since it is rotated by 45 degree, D will move on the bisector of both axis, that is D will be equidistant from two axis..
Solution
It becomes an Isosceles right angle triangle with Origin to D as the hyp.. so \(sides= hyp/\sqrt{2}=2/\sqrt{2}=\sqrt{2}\) Since D moves to III quad, x will be -ive and y positive.. ans -\(\sqrt{2},\sqrt{2}\) E
Hi Can you please elaborate the solution a bit more?
Re: A square is drawn on the xy coordinate plane as shown:
[#permalink]
Show Tags
14 Mar 2018, 19:03
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.
1) Point D is 2 units away from origin 2) the square is rotated by 45 degree..
INFERENCE
1) point D will remain 2 units away from origin.. 2) since it is rotated by 45 degree, D will move on the bisector of both axis, that is D will be equidistant from two axis..
Solution
It becomes an Isosceles right angle triangle with Origin to D as the hyp.. so \(sides= hyp/\sqrt{2}=2/\sqrt{2}=\sqrt{2}\) Since D moves to III quad, x will be -ive and y positive.. ans -\(\sqrt{2},\sqrt{2}\) E
close
63% (01:59) correct 
