Question

The incentre of the triangle whose vertices are $(1,\sqrt{3}),(0,0)$ and $(2,0)$ is

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

Answer 1

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

Let the vertices be $A(1,\sqrt{3}),B(0,0),C(2,0)$
Now,
$AB=\sqrt{(1-0{)}^2+(\sqrt{3}-0{)}^2}=2\text{units}$
$BC=\sqrt{(0-2{)}^2+(0-0{)}^2}=2\text{units}$
$CA=\sqrt{(2-1{)}^2+(0-\sqrt{3}{)}^2}=2\text{units}$

Now, incentre
$=(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c})=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})(\because a=b=c)=(\frac{1+0+2}{3},\frac{\sqrt{3}+0+0}{3})$
$=(1,\frac{1}{\sqrt{3}})$