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(A) The slopes of the lines passing through the origin are $m_1 = 1+\sqrt{2}$ and $m_2 = \frac{-1}{1+\sqrt{2}}$.Since $m_2 = \frac{-1}{1+\sqrt{2}} 	i

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$x^2+2xy-y^2=0$

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(A) The slopes of the lines passing through the origin are $m_1 = 1+\sqrt{2}$ and $m_2 = \frac{-1}{1+\sqrt{2}}$.Since $m_2 = \frac{-1}{1+\sqrt{2}} 	imes \frac{\sqrt{2}-1}{\sqrt{2}-1} = \frac{-(\sqrt{2}-1)}{2-1} = 1-\sqrt{2}$,the lines are $y = (1+\sqrt{2})x$ and $y = (1-\sqrt{2})x$.Rearranging,we get $(y - (1+\sqrt{2})x) = 0$ and $(y - (1-\sqrt{2})x) = 0$.The joint equation is $(y - (1+\sqrt{2})x)(y - (1-\sqrt{2})x) = 0$.Expanding this,we get $y^2 - (1-\sqrt{2})xy - (1+\sqrt{2})xy + (1+\sqrt{2})(1-\sqrt{2})x^2 = 0$.$y^2 - (1-\sqrt{2}+1+\sqrt{2})xy + (1-2)x^2 = 0$.$y^2 - 2xy - x^2 = 0$.Multiplying by $-1$,we get $x^2 + 2xy - y^2 = 0$.

The joint equation of lines passing through the origin and having slopes $(1+\sqrt{2})$ and $\frac{-1}{1+\sqrt{2}}$ is ...

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