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