Question

Let points $A,B,C$ lies on lines $y-x=0,2x-y=0$ and $y-3x=0$ respectively. Also $AB$ passes through fixed point $M(1,0),BC$ passes through fixed point $N(0,-1)$, then $AC$ also passes through fixed point $R(u,v)$. Find $(u+v)$.

Answer 1

Answer: 0

$\because M,A,B$ are collinear $=\left|\begin{array}{ccc}1& 0& 1\\ \alpha & \alpha & 1\\ \beta & 2\beta & 1\end{array}\right|=0\Rightarrow \alpha -2\beta +\alpha \beta =0\dots \dots$ (i)
$\because N,C,B$ are collinear
$\Rightarrow \beta -\gamma +\beta \gamma =0\Rightarrow \beta =\frac{\gamma }{1+\gamma }$ .....(ii)

Putting value of $\beta$ in equation (i), we get $\alpha =\frac{2\gamma }{1+2\gamma }$
$C,A,R$ are collinear $\Rightarrow u[\alpha -3\gamma ]-v(\alpha -\gamma )+2\alpha \gamma =0$
Put value of $\alpha$ and $\gamma$ in above equation
$\Rightarrow (-u-v)+\gamma [-6u+2v+4]=0$
$\therefore u=\frac{-1}{2},v=\frac{1}{2}$