Question

The equation of an altitude of an equilateral triangle is $\sqrt{3}x+y=2\sqrt{3}$ and one of the vertices is $(3,\sqrt{3})$, then which of the given is one of the orthocentres of the triangles?

• A. $(1,\sqrt{3}$)
• B. $(0,\sqrt{3}$)
• C. $(0,2)$
• D. none of these

Answer 1

The correct option is A $(1,\sqrt{3}$)

Refer to the figure attached.

The point $(3,\sqrt{3})$ does not satisfy the equation of the altitude $\sqrt{3}x+y=4$.

Hence it lies on the base of the equilateral triangle to which the altitude is drawn. Let $B$ be the point $(3,\sqrt{3})$ and $AD$ be the altitude.
Therefore BC is perpendicular to the altitude. Hence equation of BC is given by,

$(y-\sqrt{3})=\frac{(x-3)}{\sqrt{3}}$

$x=\sqrt{3}y$
Substituting int the equation of the altitude we get the mid point of BC as $(\frac{3}{2},\frac{\sqrt{3}}{2})$ ...(i)

Therefore If $B=(3,\sqrt{3})$ and the mid point of BC is $(\frac{3}{2},\frac{\sqrt{3}}{2})$ ...(from i)

Therefore C is $(0,0)$.
Now we know that the side AB and AC make angles of ${30}^0$ with the altitude.
Therefore considering the slope of AB or AC as $m$ and the slope of the altitude being $-\sqrt{3}$ we get,

$\tan \left({30}^0\right)=\left|\frac{m+\sqrt{3}}{1-\sqrt{3}m}\right|$

$\Rightarrow \pm \frac{1}{\sqrt{3}}=\frac{m+\sqrt{3}}{1-\sqrt{3}m}$

$\Rightarrow m=-\frac{1}{\sqrt{3}}orm=\infty$

Taking $m=-\frac{1}{\sqrt{3}}$ as slope of $AB$ equation of side $AB$ is,

$(y-\sqrt{3})=\frac{-x+3}{\sqrt{3}}$

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

$x+\sqrt{3}y=6$ ...(ii)

Solving the eqautions of the altitude of the triangle and the side AB, we get A as,

$A=(0,2\sqrt{3})$

Therefore summing up we get,
$A=(0,2\sqrt{3})$ ; $B=(3,\sqrt{3})$ ; $C=(0,0)$
Similarly, taking $m=\infty$ as slope of $AB$ we get equation of $AB$ as $x=3$,

Again solving equation of $AB$ and $AD$, we get coordinates of $A$ in this case as $(3,-\sqrt{3})$

For this case the coordinates are,

$A:(3,-3\sqrt{3})$ ; $B:(3,\sqrt{3})$ ; $C:(0,0)$

The two possible triangles are shown in the figure.

For an equilateral triangle, orthocenter and centroid coincide.

For the first triangle centroid$\equiv$Orthocenter$\equiv (\frac{3}{3},\frac{3\sqrt{3}}{3})$

Hence, $(1,\sqrt{3})$ can be the orthocenter.