If the lines \(x^2-4 x y+y^2=0\) and \(x+y=10\) contain the sides of an equilateral triangle, then the area of equilateral triangle is

\(x^2-4 x y+y^2=0\)
\(\begin{aligned} & \theta=\tan ^{-1}\left(\frac{2 \sqrt{\mathrm{h}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right) \\ & \Rightarrow \theta=\tan ^{-1}\left(\frac{2 \sqrt{2^2-1 \times 1}}{1+1}\right) \\ & \Rightarrow \theta=\tan ^{-1}(\sqrt{3})=60^{\circ}\end{aligned}\)
\(\mathrm{OM}=\frac{|0+0-10|}{\sqrt{1^2+1^2}}=5 \sqrt{2}\)
\(\tan 30^{\circ}=\frac{\mathrm{MB}}{\mathrm{OM}}=\frac{\mathrm{MB}}{5 \sqrt{2}}\)
\(\begin{aligned} & \Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{MB}}{5 \sqrt{2}} \\ & \Rightarrow \mathrm{MB}=\frac{5 \sqrt{2}}{\sqrt{3}} \\ & \Rightarrow \mathrm{AB}=2 \mathrm{MB}=\frac{10 \sqrt{2}}{\sqrt{3}}\end{aligned}\)
Now area \((\triangle \mathrm{OAB})=\frac{1}{2} \times \mathrm{AB} \times \mathrm{OM}=\frac{1}{2} \times \frac{10 \sqrt{2}}{\sqrt{3}} \times 5 \sqrt{2}=\frac{50}{\sqrt{3}}\)