The base of an equilateral triangle is represented by the equation $2 x-y-1=0$ and its vertex is $(1,2)$, then the length (in units) of the side of the triangle is

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Aicharya · Accepted Answer

$\begin{aligned} \mathrm{AD} & =\left|\frac{2-2-1}{\sqrt{2^2+(-1)^2}}ight| \ & =\left|\frac{-1}{\sqrt{5}}ight| \ & =\frac{1}{\sqrt{5}}\end{aligned}$ $\begin{aligned} & 	ext { In } 	riangle \mathrm{ABD}, 	an 60^{\circ}=\frac{\mathrm{AD}}{\mathrm{BD}} \ & \Rightarrow \sqrt{3}=\frac{\frac{1}{\sqrt{5}}}{\mathrm{BD}} \ & \Rightarrow \mathrm{BD}=\frac{1}{\sqrt{15}} \ & 	herefore \quad \mathrm{BC}=2 \mathrm{BD}=\frac{2}{\sqrt{15}}\end{aligned}$

The base of an equilateral triangle is represented by the equation $2 x-y-1=0$ and its vertex is $(1,2)$,…

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