Answer:
Option B
Explanation:
Given equations of pair of straight lines, xy-x-y+1=0 can be rewritten as
$x(y-1)-1(y-1)=0$
$\Rightarrow$ $(x-1) (y-1)=0$
$\Rightarrow$ $x-1$ and $y=1$
Thus, the three concurrent lines are
x=1 ........(i)
y=1 ......(ii)
and x+ay-3=0......(iii)
Since, Eqs.(i) and (ii) , intersect at only point , namely (1,1), therefore this point also satisfy the Eq.(iii) , we get
$1+a-3=0$
$\Rightarrow$ $a=2$
Now, the pair of lines
$ax^{2}-13xy-7y^{2}+x+23y-6=0$ becomes
$2x^{2}-13xy-7y^{2}+x+23y-6=0$
The acute angle $\theta$ between these lines is given by
$\tan \theta=\mid\frac{2\sqrt{h^{2}-ab}}{a+b}\mid=\mid\frac{2\sqrt{\left(-\frac{13}{2}\right)^{2}+14}}{-5}\mid$
$=\mid\frac{2\sqrt{\left(\frac{169+56}{4}\right)^{}}}{-5}\mid=\mid\frac{2\sqrt{\frac{225}{4}}}{-5}\mid=$
$=\mid\frac{2\times\frac{15}{2}}{-5}\mid=|-3|=3$
$\theta= tan^{1}(3)= \cos^{-1}\left(\frac{1}{\sqrt{1+3^{2}}}\right)=\cos^{-1}\left(\frac{1}{\sqrt{10}}\right)$
