Let B and C be the two points on the line y + x =0 such that B and C are symmetric with respect to the origin. Suppose A is a point on y - 2 x = 2 such that ∆ A B C is an equilateral triangle. Then, t

Question

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

TestHub · Accepted Answer

On plotting the diagram of the given data we get,

Now given, $x = y$ and $y - 2 x = 2$

On solving we get, $A (- 2, - 2)$

Now height $h$ of $∆ A B C$ $=$ Distance of point $A$ to origin,

$h = \sqrt{{(- 2)}^{2} + (- 2)^{2}}$

$\Rightarrow h = 2 \sqrt{2}$

Given $∆ A B C$ is an equilateral triangle,

So, $\sin (B) = \frac{h}{A B}$

$\Rightarrow \sin (60^{°}) = \frac{h}{A B} (∵ ∠ A = ∠ B = ∠ C = 60^{°)}$

Area of $Δ = \frac{\sqrt{3}}{4} {(A B)}^{2}$

$\Rightarrow$ Area of $Δ = \frac{\sqrt{3}}{4} \frac{h^{2}}{\sin^{2} 60^{\circ}} = \frac{\sqrt{3}}{4} \times \frac{32}{3}$

So, Area of $∆ A B C = \frac{8}{\sqrt{3}}$

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