Question

If the pair of straight lines $xy - x - y + 1 = 0$ and the line $x + a\,y - 3 = 0$ are concurrent, then the acute angle between the pair of lines $a\,x^2 - 13\,xy - 7\,y^2 + x + 23\,y - 6 = 0$ 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, $x y-x-y+1=0$ 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 $x+a y-3=0\,......(iii)$
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
$a \,x^{2}-13\,x y-7 \,y^{2}+x+23 \,y-6=0$ becomes
$2 \,x^{2}-13 \,x y-7 \,y^{2}+x+23 \,y-6=0$
The acute angle $\theta$ between these lines is given by
$\tan \theta=\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right|=\left|\frac{2 \sqrt{\left(-\frac{13}{2}\right)^{2}+14}}{-5}\right|$
$=\left|\frac{2 \sqrt{\frac{169+56}{4}}}{-5}\right|=\left|\frac{2 \sqrt{\frac{225}{4}}}{-5}\right|=\left|\frac{2 \times \frac{15}{2}}{-5}\right|=|-3|=3$
$\theta=\tan ^{-1}(3)=\cos ^{-1}\left(\frac{1}{\sqrt{1+3^{2}}}\right)=\cos ^{-1}\left(\frac{1}{\sqrt{10}}\right)$