Question

A quadrilateral ABCD is inscribed in the circle $x^2+y^2=r^2$. The slopes of AB, BC, CD are, respectively $\frac{1}{2}$, $\frac{-3}{4}$, $\frac{1}{4}$. The slope of DA is

• A. 3/2
• B. 7/2
• C. -5/2
• D. 9/2

Answer 1

The correct option is D 9/2

Given $m_{AB}=\frac{1}{2}$

$m_{BC}=-\frac{3}{4}$

$m_{CD}=-\frac{1}{4}$

Let $m_{DA}=m$

$\tan \varphi =\left|\frac{\frac{1}{4}-(-\frac{3}{4})}{1+\frac{1}{4}\left(\frac{-3}{4}\right)}\right|$

$\tan \varphi =\left|\frac{1}{1-\frac{3}{16}}\right|=\frac{16}{13}$

$\tan \theta =\left|\frac{m-\frac{1}{2}}{1+m\left(\frac{1}{2}\right)}\right|$

$\tan \theta =\left|\frac{2m-1}{2+m}\right|$

As we know opposite angles of cycle quadrilateral are supplementary :

$(\theta +\varphi )=\pi$

$\therefore \tan (\theta +\varphi )=\tan \left(\pi \right)$

$\therefore \frac{\tan \theta +\tan \varphi }{1-tan\theta \tan \varphi }=0$

$\Rightarrow \tan \theta \tan \varphi =0$

$\left(\frac{2m-1}{2+m}\right)+\left(\frac{16}{13}\right)=0$

Gives $2$ equations

$26m-13+32+16m=0$

& $26m-13-32-16m=0$

$42m+1+=0$ $10m=45$

$m=\frac{-19}{42}$ $m=\frac{45}{10}$

$m=\frac{9}{2}$

This matches one of our projects :

$\left(D\right)9/2$