Question

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

• A. $\left(1,\frac{\sqrt{3}}{2}\right)$
• B. $\left(\frac{2}{3},\frac{1}{\sqrt{3}}\right)$
• C. $\left(\frac{2}{3},\frac{\sqrt{3}}{2}\right)$
• D. $\left(1,\frac{1}{\sqrt{3}}\right)$

Answer 1

The correct option is D

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

Step 1: Find the lengths of the sides of the triangles using distance formula

Let $A=\left(1,\sqrt{3}\right)$,$B=\left(0,0\right)$, $C=\left(2,0\right)$

By using the distance formula,

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

$=2$

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

$=2$

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

$=2$

Step 2: Apply the formula for in-center of the triangle

In-center of the triangle is given by

$I=\left(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c}\right)$

$=\left(\frac{2\left(1+0+2\right)}{6},\frac{2\left(\sqrt{3}+0+0\right)}{6}\right)$

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

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