The base of an equilateral triangle is along the line given by 3x + 4y = 9. If a vertex of the triangle is (1, 2), then length of a side of the triangle is

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$\frac{4 \sqrt{3}}{5}$

$\frac{4}{5 \sqrt{3}}$

$\frac{2 \sqrt{3}}{15}$

$\frac{2 \sqrt{3}}{5}$

answer is B.

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Detailed Solution

Let AB = BC = AC = a

$\begin{array}{l} A D = ⊥ r d i s \tan c e f r o m A (1, 2) t o 3 x + 4 y - 9 = 0 \\ = \frac{|3 + 8 - 9|}{\sqrt{9 + 16}} = \frac{2}{5} \\ I n Δ^{l e} A B D, ∠ D = 90^{0} \Rightarrow B D^{2} + A D^{2} = A B^{2} \\ \Rightarrow \frac{a^{2}}{4} + \frac{4}{25} = a^{2} \\ \Rightarrow a^{2} - \frac{a^{2}}{4} = \frac{4}{25} \\ \Rightarrow \frac{3 a^{2}}{4} = \frac{4}{25} \\ \Rightarrow a^{2} = \frac{16}{3 \times 25} \Rightarrow a = \frac{4}{5 \sqrt{3}} \end{array}$