Question

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

Answer 1

$\because M , A , B$ are collinear $=\begin{vmatrix}1 & 0 & 1 \\ \alpha & \alpha & 1 \\ \beta & 2 \beta & 1\end{vmatrix}=0 \Rightarrow \alpha-2 \beta+\alpha \beta=0 \ldots \ldots$ (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[-6 u +2 v +4]=0 $
$\therefore u =\frac{-1}{2}, v =\frac{1}{2}$