The equation to the base of an equilateral triangle is x+y=2 and one vertex is (2, -1). The length of the side is

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$\sqrt{\frac{2}{3}}$

$\sqrt{\frac{3}{2}}$

$\frac{2}{\sqrt{3}}$

$\frac{1}{2 \sqrt{3}}$

answer is A.

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Detailed Solution

$\begin{array}{l} A D = ⊥ r d i s \tan c e f r o m (2, - 1) t o x + y - 2 = 0 \\ h = \frac{| 2 - 1 - 2 |}{\sqrt{1 + 1}} = \frac{1}{\sqrt{2}} \\ I n Δ^{l e} A B D \\ a^{2} = \frac{a^{2}}{4} + h^{2} \\ h^{2} = \frac{3 a^{2}}{4} \\ \Rightarrow \frac{1}{2} = \frac{3 a^{2}}{4} \\ \Rightarrow a^{2} = \frac{2}{3} \\ \Rightarrow a = \sqrt{\frac{2}{3}} \end{array}$

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