If the pair of straight lines $x y-x-y+1=0$ and the line $a x+2 y-3 a=0$ are concurrent, then $a$ is equal to
If the pair of straight lines $x y-x-y+1=0$ and the line $a x+2 y-3 a=0$ are concurrent, then $a$ is equal to
$0$
$1$
$-1$
$3$
Solution
We have,
$
\begin{array}{r}
x y-x-y+1=0 \\
a x+2 y-3 a=0
\end{array}
$
and Eqs. (i) with,
On comparing Eqs. (i) with, $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$, we get $a=0, b=0, h=\frac{1}{2}, g=-\frac{1}{2}, f=-\frac{1}{2}, c=1$
The point of concurrent is $\left(\frac{h f-b g}{a b-h^2} \frac{g h-a f}{a b-h^2}\right)$
$
=\left(\frac{\frac{1}{2}\left(-\frac{1}{2}\right)-0}{0-\left(\frac{1}{2}\right)^2,} \frac{\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)-0}{0-\left(\frac{1}{2}\right)^2}\right)=(1,1)
$
This point lies on equation (ii), then
$
\begin{array}{rlrl}
& a+2-3 a & =0 \Rightarrow 2 a=2 \\
\Rightarrow & & a & =1
\end{array}
$