Question

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

Answer 1

Step 1: Find the slope of the line and coordinates of the other vertex :

Slope of the line $y=\frac{1}{\sqrt{3}}x$ is calculated as:

$\Rightarrow {\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=30°$ with positive $x$-axis.

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

Step 2: Find the values of $x\&y$ :

Let the side be $\left(x,y\right)$.

$$\begin{gathered} \Rightarrow x^2+y^2=a^2 \\ \Rightarrow 4y^2=a^2 \\ \Rightarrow y=\pm \frac{a}{2} \end{gathered}$$

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

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