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!
 A: 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 $(1, \sqrt{3})$.
A: 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 $s=1$.  
