The joint equation of pair of lines through the origin and having slopes $(1+\sqrt{2})$ and…
The joint equation of pair of lines through the origin and having slopes $(1+\sqrt{2})$ and $\frac{1}{(1+\sqrt{2})}$ is
$x^2+2 x y+y^2=0$
$x^2-2 \sqrt{2} x y-y^2=0$
$x^2-2 \sqrt{2} x y+y^2=0$
$x^2+2 x y-y^2=0$
Solution
$\frac{1}{1+\sqrt{2}}=\frac{(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1$
Hence equations of lines passing through origin and having slopes $(\sqrt{2}+1)$ and $(\sqrt{2}-1)$ are $\mathrm{y}=(\sqrt{2}+1) \mathrm{x}$ and $\mathrm{y}=(\sqrt{2}-1) \mathrm{x}$
Their joint equation is $[(\sqrt{2}+1) \mathrm{x}-\mathrm{y}][(\sqrt{2}-1) \mathrm{x}-\mathrm{y}]=0$ $x^2-2 \sqrt{2} x y+y^2=0$