Question 397109
<font face="Times New Roman" size="+2">


Since you can translate or rotate axes any way you like, there is no loss of generality if you place your square so that one vertex is at the origin, one side coincident with the y-axis and one side coincident with the x-axis.


Let *[tex \Large a] represent the measure of a side of the square.  Then *[tex \Large (0,0),(0,a),(a,a),] and *[tex \Large (a,0)] are the coordinates of the 4 vertices.


One diagonal is contained in the line that passes through *[tex \Large (0,0)] and *[tex \Large (a,a)].  The slope of this line, call it *[tex \Large L_1] is given by the slope formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ \frac{y_1\ -\ y_2}{x_1\ -\ x_2} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_1\ =\ \frac{0\ -\ a}{0\ -\ a}\ =\ \frac{-a}{-a}\ =\ 1]


The other diagonal is contained in the line that passes through *[tex \Large (0,a)] and *[tex \Large (a,0)].  The slope of this line, call it *[tex \Large L_2] is given by the slope formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_2\ =\ \frac{a\ -\ 0}{0\ -\ a}\ =\ \frac{a}{-a}\ =\ -1]


But then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_1\ =\ -\frac{1}{m_2}]


Therefore the diagonals are perpendicular because:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  L_1\ \perp\ L_2 \ \ \Leftrightarrow\ \ m_1\ =\ -\frac{1}{m_2}\ \text{ and } m_1,\, m_2\, \neq\, 0]




John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>