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The joint equation of pair of lines through the origin and having slopes $(1+\sqrt{2})$ and $\frac{1}{(1+\sqrt{2})}$ is
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Verified Answer
The correct answer is:
$x^2-2 \sqrt{2} x y+y^2=0$
$$
\frac{1}{1+\sqrt{2}}=\frac{(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1
$$
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 $[(\sqrt{2}+1) \mathrm{x}-\mathrm{y}][(\sqrt{2}-1) \mathrm{x}-\mathrm{y}]=0$ $x^2-2 \sqrt{2} x y+y^2=0$
\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 $[(\sqrt{2}+1) \mathrm{x}-\mathrm{y}][(\sqrt{2}-1) \mathrm{x}-\mathrm{y}]=0$ $x^2-2 \sqrt{2} x y+y^2=0$
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