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