Question

The equation of the base of an equilateral triangle is $12 x+5 y-65=0$. 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 $12 x+5 y-65=0$
$\therefore A D$ is altitude of $\Delta A B C .$

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