Question

Find the incentre of the triangle whose vertices are $(2,3),(-2,-5),(-4,6)$.

Answer 1

Let

$A=\left(2,3\right)$

$B=\left(-2,-5\right)$

$C=\left(-4,6\right)$

Now by using straight line formula for two points $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$ We get,

$BC=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

$=\sqrt{{\left(-2+4\right)}^2+{\left(-5-6\right)}^2}$

$=\sqrt{2^2+{\left(-11\right)}^2}$

$=\sqrt{4+121}$

$=\sqrt{125}$

$=5\sqrt{5}$

Now , again we get

$AC=\sqrt{{\left(2+4\right)}^2+{\left(6-3\right)}^2}$

$=\sqrt{6^2+3^2}$

$=\sqrt{36+9}$

$=\sqrt{45}$

$=3\sqrt{5}$

similarly,

$AB=\sqrt{{\left(2+2\right)}^2+{\left(3+5\right)}^2}$

$=\sqrt{4^2+8^2}$

$=\sqrt{16+64}$

$=\sqrt{80}$

$=4\sqrt{5}$

Now for the incentre of the triangle we get

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

$=\left(\frac{5\sqrt{5}\times 2+3\sqrt{5}\times \left(-2\right)+4\sqrt{5}\times \left(-4\right)}{5\sqrt{5}+3\sqrt{5}+4\sqrt{5}},\frac{5\sqrt{5}\times 3+3\sqrt{5}\times \left(-5\right)+4\sqrt{5}\times 6}{5\sqrt{5}+3\sqrt{5}+4\sqrt{5}}\right)$

$=\left(\frac{10\sqrt{5}-6\sqrt{5}-6\sqrt{5}}{12\sqrt{5}},\frac{15\sqrt{5}-15\sqrt{5}+24\sqrt{5}}{12\sqrt{5}}\right)$

$=\left(\frac{-12\sqrt{5}}{12\sqrt{5}},\frac{24\sqrt{5}}{12\sqrt{5}}\right)$

$=\left(-1,2\right)$

Hence the required points of incentre is $=\left(-1,2\right)$