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 $a{x}^2-13xy-7y^2+x+23y-6=0$ is

• 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

Answer: (B)

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
$a{x}^2-13xy-7y^2+x+23y-6=0$ becomes
$2x^2-13xy-7y^2+x+23y-6=0$
The acute angle $\theta$ between these lines is given by
$\tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|=\left|\frac{2\sqrt{{\left(-\frac{13}{2}\right)}^2+14}}{-5}\right|$
$=\left|\frac{2\sqrt{\frac{169+56}{4}}}{-5}\right|=\left|\frac{2\sqrt{\frac{225}{4}}}{-5}\right|=\left|\frac{2\times \frac{15}{2}}{-5}\right|=|-3|=3$
$\theta ={\tan }^{-1}(3)={\cos }^{-1}\left(\frac{1}{\sqrt{1+3^2}}\right)={\cos }^{-1}\left(\frac{1}{\sqrt{10}}\right)$