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

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Aicharya · Accepted Answer

$x^2-4 x y+y^2=0$ $\begin{aligned} & 	heta=	an ^{-1}\left(\frac{2 \sqrt{\mathrm{h}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}ight) \ & \Rightarrow 	heta=	an ^{-1}\left(\frac{2 \sqrt{2^2-1 	imes 1}}{1+1}ight) \ & \Rightarrow 	heta=	an ^{-1}(\sqrt{3})=60^{\circ}\end{aligned}$ $\mathrm{OM}=\frac{|0+0-10|}{\sqrt{1^2+1^2}}=5 \sqrt{2}$ $	an 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 $(	riangle \mathrm{OAB})=\frac{1}{2} 	imes \mathrm{AB} 	imes \mathrm{OM}=\frac{1}{2} 	imes \frac{10 \sqrt{2}}{\sqrt{3}} 	imes 5 \sqrt{2}=\frac{50}{\sqrt{3}}$

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…

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