Question

If one vertex of an equilateral triangle of side 'a' lies at the origin and the other lies on the line $x-\sqrt{3}y=0$, then the co-ordinates of the third vertex are

• A. $(0,a)$
• B. $(\frac{\sqrt{3}a}{2},-\frac{a}{2})$
• C. $(0,-a)$
• D. $(-\frac{\sqrt{3}a}{2},\frac{a}{2})$

Answer 1

Answer: (B), (D)

The correct options are
B $(\frac{\sqrt{3}a}{2},-\frac{a}{2})$
D $(-\frac{\sqrt{3}a}{2},\frac{a}{2})$

Slope of the line $x=\sqrt{3}y$ is $\frac{1}{\sqrt{3}}$ which is ${30}^{\circ }$ with the positive x-axis.

As it is an equilateral triangle the other vertex must be on $x=-\sqrt{3}y$

Let the side be $(x,y)$. Hence,

$x^2+y^2=a^2$

$4y^2=a^2$

$y=\pm \frac{a}{2}$.

Similarily, $x=\pm \frac{\sqrt{3}a}{2}$

Hence, if the triangle lies in the 1st quadrant and 4th quadrant vertex is $(\frac{\sqrt{3}a}{2},-\frac{a}{2})$

If the triangle lies in the 2nd and 3rd quadrant the vertex is

$(-\frac{\sqrt{3}a}{2},-\frac{a}{2})$