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  • Let be the two points on the line <img alt="\mathrm{y+x=0

Let \mathrm{B \, and \: C } be the two points on the line \mathrm{y+x=0} such that \mathrm{B \: and \: C} are symmetric with respect to the origin. Suppose \mathrm{A} is a point on \mathrm{y - 2x = 2} such that \mathrm{\Delta ABC} is an equilateral triangle. Then, the area of the \mathrm{\Delta ABC} is

Option: 1

\frac{10}{\sqrt{3}}


Option: 2

3{\sqrt3}


Option: 3

2{\sqrt3}


Option: 4

\frac{8}{{\sqrt3}}


Answers (1)

best_answer

Since, A lies on perpendicular bisector of BC, whose equation is

\mathrm{y=x}                                   ...................(1)

Now, A is the point of intersection of \mathrm{y = x\: and \: y -2x = 2}

\therefore point A, after solving is \mathrm{ A(-2, -2)}

\begin{aligned} & \text { In } \triangle \mathrm{AOC} \tan 60^{\circ}=\frac{\mathrm{p}}{\mathrm{OC}} \Rightarrow \mathrm{OC}=\frac{\mathrm{p}}{\sqrt{3}}\{\because \mathrm{OA}=\mathrm{p}\} \\ & \therefore \mathrm{BC}=2 \times \mathrm{OC}=\frac{2 \mathrm{p}}{\sqrt{3}} \end{aligned}
Now, Area of \triangle \mathrm{ABC}=\frac{1}{2} \times \mathrm{BC} \times \mathrm{OA}
                                  =\frac{1}{2} \times \frac{2 \mathrm{p}}{\sqrt{3}} \times \mathrm{p}=\frac{\mathrm{p}^2}{\sqrt{3}} \text { sq. unit }
and \mathrm{ \mathrm{p}=\mathrm{OA}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}}
So, Area of \mathrm{\triangle \mathrm{ABC}=\frac{(2 \sqrt{2})^2}{\sqrt{3}}=\frac{8}{\sqrt{3}} sq. unit}

Posted by

Pankaj Sanodiya

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