Question

The equation of an altitude of an equilateral triangle is $\sqrt{3}x+y=2\sqrt{3}$ and one of its vertices is $(3,\sqrt{3})$ then, the possible number of triangle is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$2$

REF.Image

Equation of $AD$ is

$\sqrt{3}x+y=2\sqrt{3}$

Hence, maximum 'two'

equilateral can be drawn.