The acute angle between the pair of straight | vedclass

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(C) Given the equation of the pair of lines is $k x^2+8 x y-3 y^2+2 x-4 y-1=0$.For this to represent a pair of straight lines,the determinant must be

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$\cos ^{-1}\left(\frac{1}{\sqrt{10}}\right)$

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(C) Given the equation of the pair of lines is $k x^2+8 x y-3 y^2+2 x-4 y-1=0$.For this to represent a pair of straight lines,the determinant must be zero:$\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$.Now,homogenize the equation $3 x^2+8 x y-3 y^2+2 x-4 y-1=0$ using $x+y=1$:$3 x^2+8 x y-3 y^2+(2 x-4 y)(x+y) - (x+y)^2 = 0$$3 x^2+8 x y-3 y^2+2 x^2+2 x y-4 x y-4 y^2 - (x^2+2 x y+y^2) = 0$$4 x^2+4 x y-8 y^2 = 0 \Rightarrow x^2+x y-2 y^2 = 0$.Comparing with $A x^2+2 H x y+B y^2=0$,we have $A=1, H=1/2, B=-2$.The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2 \sqrt{H^2-A B}}{A+B} ight|$:$	an 	heta = \left| \frac{2 \sqrt{(1/2)^2 - (1)(-2)}}{1-2} ight| = \left| \frac{2 \sqrt{1/4+2}}{-1} ight| = 2 \sqrt{9/4} = 2 	imes \frac{3}{2} = 3$.Since $	an 	heta = 3$,we have $\cos 	heta = \frac{1}{\sqrt{1+3^2}} = \frac{1}{\sqrt{10}}$.Therefore,$	heta = \cos ^{-1}\left(\frac{1}{\sqrt{10}}ight)$.

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

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