Question

The acute angle between the pair of straight lines joining the origin to the points of intersection of the line $x+y-1=0$ with the pair of straight lines $k x^{2}+8 x y-3 y^{2}+2 x-4 y-1=0$ is

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{4}$
• C. $\cos^{-1}\left(\frac{1}{\sqrt{10}}\right)$
• D. $\cos^{-1}\left(\frac{3}{\sqrt{2}}\right)$

Answer 1

Answer: (C)

We have
$k x^{2}+8 x y-3 y^{2}+2 x-4 y-1=0$ is form of a straight line
$\therefore \begin{vmatrix}k & 4 & 1 \\ 4 & -3 & -2 \\ 1 & -2 & -1\end{vmatrix}=0$
$k(3-4)-4(-4+2)+1(-8+3)=0$
$-k+8-5=0 $
$\Rightarrow k=3$
$\therefore $ Acute angle between line joining the origin $\alpha$ the point of intersection $x+y=1$ and line
$3 x^{2}+8 x y-3 y^{2}+2 x-4 y-1=0$ by
Homogenous using the equation
$\therefore 3 x^{2}+8 x y-3 y^{2}+2 x(x+y)-4 y(x+y)-(x+y)^{2}=0 $
$3 x^{2}+8 x y-3 y^{2}+2 x^{2}+2 x y-4 x y-4 y^{2}-x^{2}-y^{2}-2 x y=0 $
$4 x^{2}+4 x y-8 y^{2}=0 $
$\Rightarrow x^{2}+x y-2 y^{2}=0$
$\therefore \tan \theta=\left|\frac{2 \sqrt{(1 / 2)^{2}+2}}{1-2}\right|=3 $
$\Rightarrow \cos \theta=\frac{1}{\sqrt{1+3^{2}}}$
$\cos \theta=\frac{1}{\sqrt{10}} $
$\Rightarrow \theta=\cos ^{-1}\left(\frac{1}{\sqrt{10}}\right)$