The joint equation of the pair of lines passi | vedclass

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(C) The slopes of the two lines are $m_1 = 1+\sqrt{2}$ and $m_2 = \frac{1}{1+\sqrt{2}}$.Rationalizing $m_2$: $m_2 = \frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sq

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$x^2-2\sqrt{2}xy+y^2=0$

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(C) The slopes of the two lines are $m_1 = 1+\sqrt{2}$ and $m_2 = \frac{1}{1+\sqrt{2}}$.Rationalizing $m_2$: $m_2 = \frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)} = \frac{\sqrt{2}-1}{2-1} = \sqrt{2}-1$.The equations of the lines passing through the origin are $y = m_1x$ and $y = m_2x$,which are $y = (1+\sqrt{2})x$ and $y = (\sqrt{2}-1)x$.Rearranging these gives $(1+\sqrt{2})x - y = 0$ and $(\sqrt{2}-1)x - y = 0$.The joint equation is given by the product: $[(1+\sqrt{2})x - y][(\sqrt{2}-1)x - y] = 0$.Expanding this: $(1+\sqrt{2})(\sqrt{2}-1)x^2 - (1+\sqrt{2})xy - (\sqrt{2}-1)xy + y^2 = 0$.$(2-1)x^2 - (1+\sqrt{2}+\sqrt{2}-1)xy + y^2 = 0$.$x^2 - 2\sqrt{2}xy + y^2 = 0$.

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

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