Question

The in-centre of a triangle with vertices $\left(7,1\right)$, $\left(-1,5\right)$and $\left(3+2\sqrt{3},3+4\sqrt{3}\right)$ is

• A. $\left(3+\frac{2}{\sqrt{3}},3+\frac{4}{\sqrt{3}}\right)$
• B. $\left(1+\frac{2}{3\sqrt{3}},1+\frac{4}{3\sqrt{3}}\right)$
• C. $\left(7,1\right)$
• D. None of these

Answer 1

The correct option is A

$\left(3+\frac{2}{\sqrt{3}},3+\frac{4}{\sqrt{3}}\right)$

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

Let $A=\left(7,1\right)$, $B=\left(-1,5\right)$, $C=\left(3+2\sqrt{3},3+4\sqrt{3}\right)$

$AB\left(c\right)=\sqrt{{\left(7+1\right)}^2+{\left(1-5\right)}^2}=\sqrt{80}$

$=4\sqrt{5}$

$BC\left(a\right)=\sqrt{{\left(-1-3-2\sqrt{3}\right)}^2+{\left(5-3-4\sqrt{3}\right)}^2}=\sqrt{16+12+4+48}$

$=4\sqrt{5}$

$AC\left(b\right)=\sqrt{{\left(7-3-2\sqrt{3}\right)}^2+{\left(1-3-4\sqrt{3}\right)}^2}=\sqrt{16+12+4+48}$

$=4\sqrt{5}$

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

The incenter of the triangle is given as

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

$=\left(\frac{4\sqrt{5}\left(7-1+3+2\sqrt{3}\right)}{12\sqrt{5}},\frac{4\sqrt{5}\left(1+5+3+4\sqrt{3}\right)}{12\sqrt{5}}\right)$

$=\left(\frac{9+2\sqrt{3}}{3},\frac{9+4\sqrt{3}}{3}\right)$

$I=\left(3+\frac{2}{\sqrt{3}},3+\frac{4}{\sqrt{3}}\right)$

Hence, the in-centre of the triangle is $\left(3+\frac{2}{\sqrt{3}},3+\frac{4}{\sqrt{3}}\right)$, so, option $\left(A\right)$ is the correct answer.