Question

The incenter of the triangle with vertices $(1,\sqrt{3}),(0,0),\text{and}(2,0)$ is

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

Answer 1

The correct option is D $(1,1/\sqrt{3})$

$LettheverticesofA\left(1,\sqrt{3}\right),B\left(0,0\right)andC\left(2,0\right)again$

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

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

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

since, all sides of a triangle are equal then the triangle is an equilateral triangle

Hence, coordinates of incentre are

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

Hence, the option $D$ is the correct answer.