Question

If the pair of straight lines $xy-x-y+1=0$ and the line $x+ay-3=0$ are concurrent, then the acute angle between the pair of lines $a{x}^2-13xy-7y^2+x+23y-6=0$

• A. ${\cos }^{-1}\left(\frac{5}{\sqrt{218}}\right)$
• B. ${\cos }^{-1}\left(\frac{1}{\sqrt{10}}\right)$
• C. ${\cos }^{-1}\left(\frac{5}{\sqrt{173}}\right)$
• D. ${\cos }^{-1}\left(\frac{1}{\sqrt{5}}\right)$

Answer 1

The correct option is B ${\cos }^{-1}\left(\frac{1}{\sqrt{10}}\right)$

$xy-x-y+1+0$ ....(1)
$x+ay-3=0\to \left(2\right)$
$\Rightarrow$ $x(y-1)-(y-1)=0$
$\Rightarrow$ $(x-1)(y-1)=0$
Point of intersection $(1,1)$
(1) (2) are concurrent $\Rightarrow$ passing through $(1,1)$
Therefore, $1+a-3=0\Rightarrow$ $a=2$
Now $2x^2-13xy-7y^2+x+23y-6=0$
$\cos \theta =\frac{|a+b|}{\sqrt{{(a-b)}^2+4h^2}}=\frac{5}{250}=\frac{1}{\sqrt{10}}\Rightarrow \theta ={\cos }^{-1}\frac{1}{\sqrt{10}}$