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Question Number 50676 by Tinkutara last updated on 18/Dec/18

Commented by ajfour last updated on 18/Dec/18

Commented by ajfour last updated on 18/Dec/18

$$E(\frac{a}{2}\sin {}^2\theta ,\frac{a}{2}\cos \theta \sin \theta )$$$$m_{BE}=\frac{\sin \theta \cos \theta }{1+\sin {}^2\theta }$$$$m_{AF}=-\frac{\tan \theta -\frac{\sin \theta \cos \theta }{2}}{\frac{\sin {}^2\theta }{2}}$$$$=-\frac{2-\cos {}^2\theta }{\sin \theta \cos \theta }=-\frac{1}{m_{BE}}$$$$henceAF\mathrm{\perp }BE.$$

Commented by Tinkutara last updated on 19/Dec/18

Thanks Sir! Excellent method, simple and short!

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