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 the possible number of triangles are,

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

Let ABC be the triangle, where BC is the base of a triangle as given in the figure.

If we put the vertices $(3,\sqrt{3})$ in the given equation, then vertices do not satisfy the equation of the altitude.

Hence it lies on the base of the equilateral triangle to which the altitude is drawn.

Now, 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 makes an angle of ${30}^0$ with the altitude.
Therefore considering the slope of AB as $m$ and the slope of the altitude being $-\sqrt{3}$ we get,

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

Therefore

$1-\sqrt{3}m=\sqrt{3}m+3$

$m=\frac{-1}{\sqrt{3}}$

Hence 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 equations of the altitude of the triangle and the side AB, we get A as
$A=(0,2\sqrt{3})$

Therefore summing up we get

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

However if we take

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

We do not obtain any values of $m.$ This is due the fact that another vertex may lie on the altitude with $x=3$
Substituting in the equation of the altitude, $x=3$, we get
$y=-\sqrt{3}$
Hence another vertex is $(3,\sqrt{3})$

Hence there will be a total of two equilateral triangles with vertices

$A=(0,2\sqrt{3})$ ; $B=(3,\sqrt{3})$ ; $C=(0,0)$
and
$A=(3,-\sqrt{3})$ ; $B=(3,\sqrt{3})$ ; $C=(0,0)$