Question

The equation of the base of an equilateral triangle is $12x+5y-65=0$. If one of its vertices is $(2,3)$, then the length of the side is

• A. $\frac{4}{13}$
• B. $\frac{2}{\sqrt{3}}$
• C. $\frac{4}{\sqrt{3}}$
• D. $\frac{2}{13}$

Answer 1

Answer: (C)

Since, $A(2,3)$ do not lie on $12x+5y-65=0$
$\therefore AD$ is altitude of $\Delta ABC.$

$\therefore$ Length of $AD=\left|\frac{12\times 2+5\times 3-65}{\sqrt{12^2+5^2}}\right|$
$=\left|\frac{24+15-65}{13}\right|=\left|\frac{26}{13}\right|=2$
We know that Length of altitude $=\frac{\sqrt{3}}{2}\times$ side
$\therefore$ Side $=\frac{2}{\sqrt{3}}\times AD$
$=\frac{2}{\sqrt{3}}\times 2=\frac{4}{\sqrt{3}}$