The line x² + 4xy + y² = 0 and x - y = 4 are the sides of an equilateral triangle whose area is equal to A/(3)^1/2 , then A is equal to? 
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$$\begin{gathered} \mathrm{Let} \mathrm{equilateral} \mathrm{triangle} \mathrm{be} \mathrm{ABC} \\ {\mathrm{x}}^2+4\mathrm{xy}+{\mathrm{y}}^2=0 ;\left(\mathrm{i}\right) \\ \mathrm{Above} \mathrm{equation} \mathrm{represents} \mathrm{pair} \mathrm{of} \mathrm{straight} \mathrm{lines} \mathrm{passing} \mathrm{through} \mathrm{origin}, \mathrm{hence} \\ \mathrm{A}\left(0,0\right) \mathrm{be} \mathrm{one} \mathrm{of} \mathrm{the} \mathrm{vertices}. \\ \mathrm{x}-\mathrm{y}=4 \\ \Rightarrow \mathrm{x}=\mathrm{y}+4 \\ \mathrm{Put} \mathrm{value} \mathrm{of} \mathrm{x} \mathrm{in} \left(\mathrm{i}\right) \\ {\left(\mathrm{y}+4\right)}^2+4\left(\mathrm{y}+4\right)\mathrm{y}+{\mathrm{y}}^2=0 \\ \Rightarrow {\mathrm{y}}^2+16+8\mathrm{y}+4{\mathrm{y}}^2+16\mathrm{y}+{\mathrm{y}}^2=0 \\ \Rightarrow 6{\mathrm{y}}^2+24\mathrm{y}+16=0 \\ \Rightarrow 3{\mathrm{y}}^2+12\mathrm{y}+8=0 \\ \mathrm{By} \mathrm{Sridharacharya}\text{'}\mathrm{s} \mathrm{formula}, \mathrm{we} \mathrm{get}: \\ \mathrm{y}=\frac{-12\pm \sqrt{{12}^2-4\times 3\times 8}}{2\times 3}=\frac{-12\pm \sqrt{48}}{6} \\ =\frac{-12\pm 4\sqrt{3}}{6} \\ =-2\pm \frac{2}{3}\sqrt{3} \\ \mathrm{Hence} \mathrm{y}=-2+\frac{2}{3}\sqrt{3} \mathrm{or} \mathrm{y}=-2-\frac{2}{3}\sqrt{3} \\ \mathrm{When} \mathrm{y}=-2+\frac{2}{3}\sqrt{3} \\ \mathrm{x}=\mathrm{y}+4=-2+\frac{2}{3}\sqrt{3}+4=2+\frac{2}{3}\sqrt{3} \\ \mathrm{Hence} \mathrm{coordinates} \mathrm{of} \mathrm{B} \mathrm{will} \mathrm{be} \mathrm{B}\left(2+\frac{2}{3}\sqrt{3},-2+\frac{2}{3}\sqrt{3}\right) \\ \mathrm{Side} \mathrm{of} \mathrm{equilateral} \mathrm{triangle}\left(\mathrm{a}\right)=\mathrm{AB}=\sqrt{{\left(2+\frac{2}{3}\sqrt{3}-0\right)}^2+{\left(-2+\frac{2}{3}\sqrt{3}-0\right)}^2} \\ =\sqrt{{\left(2+\frac{2}{3}\sqrt{3}\right)}^2+{\left(-2+\frac{2}{3}\sqrt{3}\right)}^2} \\ =\sqrt{2^2+{\left(\frac{2}{3}\sqrt{3}\right)}^2+2.2.\frac{2}{3}\sqrt{3}+2^2+{\left(\frac{2}{3}\sqrt{3}\right)}^2-2.2.\frac{2}{3}\sqrt{3}} \\ =\sqrt{4+\frac{4}{3}+4+\frac{4}{3}} \\ =\sqrt{8+\frac{8}{3}}=\sqrt{\frac{32}{3}}=4\sqrt{\frac{2}{3}}=\mathrm{a} \\ \mathrm{Area} \mathrm{of} \mathrm{equilateral} \mathrm{triangle}=\frac{\sqrt{3}}{4}{\mathrm{a}}^2 \\ =\frac{\sqrt{3}}{4}{\left(4\sqrt{\frac{2}{3}}\right)}^2 \\ =\frac{\sqrt{3}}{4}\times 16\times \frac{2}{3} \\ =\frac{8}{\sqrt{3}} \\ \mathrm{According} \mathrm{to} \mathrm{ques} \\ \frac{8}{\sqrt{3}}=\frac{\mathrm{A}}{\sqrt{3}} \\ \mathbf{Hence}\mathbf{ }\mathbf{A}\mathbf{=}\mathbf{8}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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