urgent
​The centroid of an equi. triangle is (0,0) . If two vertices of the triangle lie on x+y = 2root2 then one of them will have coordinates as 
Ans: (root2+root6. root2-root6)

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$$\begin{gathered} \mathrm{Perpendicular} \mathrm{distance} \mathrm{of} \left(0,0\right) \mathrm{from} \mathrm{x}+\mathrm{y}-2\sqrt{2}=0 \mathrm{is} \mathrm{given} \mathrm{by}: \\ \mathrm{p}=\frac{\left|0+0-2\sqrt{2}\right|}{\sqrt{1^2+1^2}}=\frac{\left|-2\sqrt{2}\right|}{\sqrt{2}}=\frac{2\sqrt{2}}{\sqrt{2}}=2 \\ \mathrm{For} \mathrm{equilateral} \mathrm{triangle}, \mathrm{if} \mathrm{height} \mathrm{is} \mathrm{h} \mathrm{then}: \\ \frac{\mathrm{h}}{3}=\mathrm{p}=2 \\ \Rightarrow \mathrm{h}=6 \\ \mathrm{Let} \mathrm{length} \mathrm{of} \mathrm{side} \mathrm{be} \text{'}\mathrm{a}\text{'} \\ \mathrm{Area}=\mathrm{Area} \\ \Rightarrow \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}=\frac{\sqrt{3}}{4} {\mathrm{a}}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{a}\times 6=\frac{\sqrt{3}}{4} {\mathrm{a}}^2 \\ \Rightarrow \mathrm{a}=\frac{12}{\sqrt{3}}=4\sqrt{3} \\ \Rightarrow \frac{\mathrm{a}}{2}=2\sqrt{3} \\ \mathrm{CD}=\frac{\mathrm{a}}{2}=2\sqrt{3} \\ \mathrm{OD}=\mathrm{p}=2 \\ \mathrm{By} \mathrm{Pythagorus} \mathrm{theorem} \\ \mathrm{OC}=\sqrt{{\mathrm{OD}}^2+{\mathrm{CD}}^2}=\sqrt{2^2+{\left(2\sqrt{3}\right)}^2}=\sqrt{4+12}=4 \\ \mathrm{Any} \mathrm{point} \mathrm{on} \mathrm{line} \mathrm{x}+\mathrm{y}-2\sqrt{2}=0 \mathrm{will} \mathrm{be} \left(\mathrm{h},2\sqrt{2}-\mathrm{h}\right) \\ \mathrm{OC}=4 \\ \sqrt{{\left(\mathrm{h}-0\right)}^2+{\left(2\sqrt{2}-\mathrm{h}\right)}^2}=4 \\ \Rightarrow {\mathrm{h}}^2+8+{\mathrm{h}}^2-4\sqrt{2}\mathrm{h}=16 \\ \Rightarrow 2{\mathrm{h}}^2-4\sqrt{2}\mathrm{h}+8-16=0 \\ \Rightarrow 2{\mathrm{h}}^2-4\sqrt{2}\mathrm{h}-8=0 \\ \Rightarrow {\mathrm{h}}^2-2\sqrt{2}\mathrm{h}-4=0 \\ \mathrm{Using} \mathrm{Shridaracharya}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{get}: \\ \mathrm{h}=\frac{-\left(-2\sqrt{2}\right)\pm \sqrt{{\left(2\sqrt{2}\right)}^2-4\times 1\times \left(-4\right)}}{2} \\ =\frac{2\sqrt{2}\pm \sqrt{8+16}}{2} \\ =\frac{2\sqrt{2}\pm \sqrt{24}}{2} \\ =\frac{2\sqrt{2}\pm 2\sqrt{6}}{2} \\ =\sqrt{2}\pm \sqrt{6} \\ \mathrm{When} \mathrm{we} \mathrm{take} \mathrm{positive} \mathrm{sign} \\ \mathrm{h}=\sqrt{2}+\sqrt{6} \\ 2\sqrt{2}-\mathrm{h}=2\sqrt{2}-\sqrt{2}-\sqrt{6}=\sqrt{2}-\sqrt{6} \\ \mathbf{Hence}\mathbf{ }\mathbf{one}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{vertex}\mathbf{ }\mathbf{is}\mathbf{ }\left(\sqrt{\mathbf{2}}\mathbf{+}\sqrt{\mathbf{6}}\mathbf{,}\sqrt{\mathbf{2}}\mathbf{-}\sqrt{\mathbf{6}}\right) \end{gathered}$$

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