Question

The vertex of an equilateral triangle is $(2,-1)$ and the equation of its base is $x+2y=1$. The length of its sides is

• A. $\frac{4}{\sqrt{15}}$
• B. $\frac{2}{\sqrt{15}}$
• C. $\frac{4}{3\sqrt{3}}$
• D. $\frac{1}{\sqrt{5}}$

Answer 1

The correct option is B $\frac{2}{\sqrt{15}}$

p = length of perpendicular from A $(2,-1)$ to BC.
$p=\left|\frac{2+2(-1)-1)}{\sqrt{5}}\right|=\frac{1}{\sqrt{5}}$
Now In $\Delta ABD$
$\sin {60}^{\circ }=\frac{p}{AB}$
$\frac{\sqrt{3}}{2}=\frac{1}{\sqrt{5}}\frac{1}{AB}$
$AB=\frac{2}{\sqrt{15}}$