# rotating 90 degrees around a circle on a co-ordinate plane

I thought the answer would be square root of 3. It would seem that the x co-ordinate of Q would just be the opposite of the x co-ordinate of P.

I'm not sure if the picture is just being deceptive, or if I just don't remember my math from high school very well...

However... I'm told the correct answer is 1.

I don't understand why... Could someone explain that please?

Thanks so much!

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Someone is sure to leave a great answer shortly, so I'm just gonna drop this comment and say that when you rotate by 90 degrees, you swap the x and y coordinates and then negate one of them, depending on which direction you rotated. –  El'endia Starman Aug 16 '11 at 1:26
@El'endia Starman: Seems like a good clear answer to me, with the addition of a little detail. –  André Nicolas Aug 16 '11 at 1:29
it just seems to me that along the x axis, the point P is as far from the origin as the point Q... But maybe just because the picture is not drawn to scale? –  kralco626 Aug 16 '11 at 1:29
That is indeed the case; the picture is not quite correct. That tends to happen with most images on tests/homework, actually... –  El'endia Starman Aug 16 '11 at 1:31
it's actually a question on an "official" GMAT preparation test, for which the directions say "All figures are drawn as accurately as possible. Exceptions will be clearly noted"... but anyways I now understand and will hopefully get it right on the actual test! –  kralco626 Aug 16 '11 at 1:35

A simple answer to your question is that when you rotate by $90$ degrees (as indicated by the right angle symbol), you swap the $x$ and $y$ coordinates and then negate one or the other, depending on which direction you rotated it. In your case, you had $(-\sqrt{3}, 1)$, which became $(1, -\sqrt{3})$ and then because you rotated into the first quadrant, the final point is $(1, \sqrt{3})$.

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Thanks for the quick reply! –  kralco626 Aug 16 '11 at 1:39
@kralco626: You're welcome. :) –  El'endia Starman Aug 16 '11 at 1:41

There are various "high school" approaches to the answer. You will have to help me by drawing a picture.

Drop a perpendicular from the point $P$ to a point $M$ on the negative $x$-axis. Look at the angle $MOP$, and call it $\theta$. In $\triangle OPM$, the hypotenuse $OP$ has length $\sqrt{(\sqrt{3})^2+1}$, which is $2$. Thus $\sin\theta=1/2$.

You may recall "special angles." The angle $\theta$ between $0^\circ$ and $90^\circ$ such that $\sin\theta=1/2$ is the $30^\circ$ angle.

Now drop a perpendicular from $Q$ to the point $N$ on the positive $x$-axis. Let $\phi$ be the angle $QON$. What is the size of $\phi$? It is $180^\circ-(90^\circ+30^\circ)$, which is $60^\circ$. Thus $\phi$ is a lot bigger than the $30^\circ$ angle $\theta$, so the picture should not be at all symmetrical bout the $y$-axis!

Note that the cosine of the angle $\phi$ is $s/2$. But the cosine of the $60^\circ$ angle is $1/2$. It follows that $s=1$.

Without special angles: Do the constructions of $M$ and $N$ exactly as in the first solution, and let $\theta$, $\phi$ be as described there.

Note that $\theta+\phi=90^\circ$, so $\theta$ and $\phi$ are complementary angles.

Now compare $\triangle OPM$ and $\triangle QON$. We have $OP=QO$, and $\angle OPM=\angle QON$. So the triangles are congruent. Note that $PM=ON$. But $PM=1$, and therefore $s=ON=1$.

Using some analytic geometry: The slope of the line $OP$ is $-1/\sqrt{3}$. But $OQ$ is perpendicular to $OP$, so its slope is the negative reciprocal of $-1/\sqrt{3}$. Thus the slope of $OQ$ is $\sqrt{3}/1$.

But the slope of $OQ$ is $t/s$. It follows that $t=s\sqrt{3}$. By the Pythagorean Theorem, $s^2+t^2=4$. So $s^2+3s^2=4$. Since $s$ is positive, we conclude that $s=1$.

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