Question

The equation of an altitude of an equilateral triangle is $\sqrt{3}x+y=2\sqrt{3}$ and one of the vertex is $(3,\sqrt{3})$. Then

• A. The possible number of triangles is 2
• B. Orthocentre of one of the triangle is $(1,\sqrt{3})$
  
• C. Area of the triangle is $3\sqrt{3}$ sq. units.
  
• D. The possible vertices are $(0,2\sqrt{3}),(3,-\sqrt{3})$ and (0, 0)

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A The possible number of triangles is 2
B Orthocentre of one of the triangle is $(1,\sqrt{3})$

C Area of the triangle is $3\sqrt{3}$ sq. units.

D The possible vertices are $(0,2\sqrt{3}),(3,-\sqrt{3})$ and (0, 0)

One side is $x=\sqrt{3y}$
Image of $(3,\sqrt{3})$ w.r.t altitude is (0,0)
Other vertex may be $(0,2\sqrt{3})$ or $(3,-\sqrt{3})$
side=$2\sqrt{3}$
G=(1, \sqrt{3})