Question

Write the coordinates of the incentre of the triangle having its vertices at(0,0),(5,0) and (0,12).

Answer 1

$\text{Let the given vertices of the triangle are}A=(x_1,y_1)=(0,0)B=(x_2,y_2)=(5,0)C=(x_3,y_3)=(0,12)Now,a=BC=\sqrt{(0-5{)}^2+(12-0{)}^2}=\sqrt{25+144}=\sqrt{169}=13b=AC=\sqrt{(0-0{)}^2+(12-0{)}^2}=12andcAB\sqrt{(5-0{)}^2+(0-0{)}^2}=5\text{The coordinates of the in-centre of the triangle ABC are}(\frac{a{x}_1+b{x}_2+cx+3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c})or(\frac{13\times 0+12\times 5+5\times 0}{13+12+5},\frac{13\times 0+12\times 0+5\times 12}{13+12+15})or(\frac{60}{30},\frac{60}{30})or(2,2)\text{Hence,the coordinates of the in centre are (2,2)}$