Question

The in-centre of a triangle with vertices $(1,\sqrt{3})$, $(0,0)$, and $(2,0)$ is

• A. $\left(1,\frac{1}{\sqrt{3}}\right)$
• B. $\left(1,\frac{2}{\sqrt{3}}\right)$
• C. $\left(1,\frac{\sqrt{3}}{2}\right)$
• D. None of these

Answer 1

The correct option is A

$\left(1,\frac{1}{\sqrt{3}}\right)$

Step 1: Using distance formula, find the lengths of the sides of the triangle

Let $A(1,\sqrt{3})$, $B(0,0)$, and $C(2,0)$ be the vertices of the triangle.

Using distance formula,

$AB=\sqrt{{\left(1-0\right)}^2+{\left(\sqrt{3}-0\right)}^2}$

$=2$ $...\left(i\right)$

$BC=\sqrt{{\left(0-2\right)}^2+{\left(0-0\right)}^2}$

$=2$ $...\left(ii\right)$

$AC=\sqrt{{\left(1-2\right)}^2+{\left(\sqrt{3}-0\right)}^2}$

$=2$ $...\left(iii\right)$

Step 2: Deduce the type of the triangle

From $\left(i\right),\left(ii\right),\left(iii\right)$ we get that

$AB=AC=BC$

$\Rightarrow$$△ABC$ is an equilateral triangle

In an equilateral triangle, the centroid and in-centre are coincident.

Step 3: Use formula for centroid of the triangle

The centroid of the triangle is given as

$G=\left(\frac{x_A+x_B+x_C}{3},\frac{y_A+y_B+y_C}{3}\right)$

$=\left(\frac{1+0+2}{3},\frac{\sqrt{3}+0+0}{3}\right)$

$\therefore$ $G=\left(1,\frac{1}{\sqrt{3}}\right)$

Hence, the in-centre of the triangle is $\left(1,\frac{1}{\sqrt{3}}\right)$, so, option $\left(A\right)$ is the correct answer.