Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

Let the triangle be $A B C$ with $C \equiv(3, \sqrt{3})$ and the altitude drawn through the vertex (meeting $B C$ at $D$ ) be $\sqrt{3} x+ y-2 \sqrt{3}=0$.
If $B$ is $\left(x_{b}, y_{b}\right)$, then we have
$\frac{x_{b}-3}{\sqrt{3}}=\frac{y_{b}-\sqrt{3}}{1}=\frac{2(3 \sqrt{3}+\sqrt{3}-2 \sqrt{3})}{4}=-\sqrt{3}$
or $ x_{b}=0, y_{b}=0$
and the coordinates of $D$ are $(3 / 2, \sqrt{3} / 2)$.
Let the coordinates of vertex $A$ be $\left(x_{a}, y_{a}\right)$. Then,
$\frac{x_{a}-(3 / 2)}{-1 / 2}=\frac{y_{a}-(\sqrt{3} / 2)}{\sqrt{3} / 2}=\pm 3$
or $\left(x_{a}, y_{a}\right) \equiv(0,2 \sqrt{3})$ or $(3,-\sqrt{3})$
Hence, the remaining vertices are $(0,0)$ and $(0,2 \sqrt{3})$
or $(0,0)$ and $(3,-\sqrt{3})$.
Also, the orthocenter is $(1, \sqrt{3})$
or $(2,0)$.