The circumcentre of the triangle formed by the points $A(1, \sqrt{3}), B(-1,-\sqrt{3})$ and $(3,-\sqrt{3})$ is

The circumcentre of the triangle formed by the points $A(1, \sqrt{3}), B(-1,-\sqrt{3})$ and $(3,-\sqrt{3})$ is
  1. $(1,-\sqrt{3})$
  2. $\left(-1, \frac{1}{\sqrt{3}}\right)$
  3. $(0,0)$
  4. $\left(1, \frac{-1}{\sqrt{3}}\right)$

Solution

Vertices of $\triangle A B C$ are $A(1, \sqrt{3}), B(-1,-\sqrt{3})$ and $C(3,-\sqrt{3})$ $A B=\sqrt{4+12}=4$ $\begin{aligned} & B C=\sqrt{4+12}=4 \\ & A C=\sqrt{4+12}=4\end{aligned}$ Here, $A B C$ is an equilateral triangle. $\therefore$ Circumcentre of triangle is $\left(\frac{1-1+3}{3}, \frac{\sqrt{3}-\sqrt{3}-\sqrt{3}}{3}\right)$ $=\left(1, \frac{-1}{\sqrt{3}}\right)$ $[\because$ in equilateral triangle circumcentre and centroid coincide]

Asked in: AP EAMCET 2021 (24 Aug Shift 2)

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