If the angle between the pair of lines $x^2+2\sqrt{2}xy+k{y}^2=0,k>0is{45}^0,$ then the area ( in square units ) of the triangle formed by the pair of bisectors of angles between the given lines and the line x + 2y +1 =0 is

Given Equation $x^2+2\sqrt{2}xy+k{y}^2=0,\theta ={45}^0,$

$\begin{array}{l}\cos {45}^0=\frac{\left|1+k\right|}{\sqrt{{\left(1-k\right)}^2+8}}\\ \Rightarrow \frac{1}{\sqrt{2}}=\frac{\left|k+1\right|}{\sqrt{{\left(1-k\right)}^2+8}}\\ \Rightarrow {\left(1-k\right)}^2+8=2{\left(k+1\right)}^2\\ \Rightarrow 1+k^2-2k+8=2k^2+2+4k\\ \Rightarrow k^2+6k-7=0\\ \Rightarrow k^2+7k-k-7=0\\ \Rightarrow \left(k+7\right)\left(k-1\right)=0\\ k=1,k=-7\\ \because k>0\Rightarrow k=1\end{array}$               

The Equation is  $x^2+2\sqrt{2}xy+y^2=0$

Pair of angle bisectors

  $\begin{array}{l}\sqrt{2}\left(x^2-y^2\right)=\left(1-1\right)xy\\ \Rightarrow x^2-y^2=0\end{array}$             

Comparing with

$\begin{array}{l}a{x}^2+2hxy+b{y}^2=0\\ a=1,h=0,b=-1\\ andx+2y+1=0comparingwithlx+my+n=0\\ l=1,m=2,n=1\\ Area=\frac{h^2\sqrt{h^2-ab}}{\left|a{m}^2-2hlm+b{l}^2\right|}=\frac{1\sqrt{0+1}}{\left|1\left(4\right)-0-1\left(1\right)\right|}=\frac{1}{3}\end{array}$