Question

$\text{Find the coordinates of the centre of the circle inscribed in a triangle whose vertices are A(-36,7),B(20,7)and C(0,-8).}$

Answer 1

$\text{The coordinates of the in-centre whose vertices are}A(x_1,y_1),B(x_x,y_2),andC(x_3,y_3)are\left[\begin{array}{c}\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c}\end{array}\right],Wherea=BC,b=ACandc=ABLetA(-36,7)B(20,7)and(0,-8)betheverticesofthetrangleABCNow,a=BC=\sqrt{(0-20{)}^2+(-8-7{)}^2}=\sqrt{400+225}=\sqrt{625}=25,b=AC=\sqrt{(0+36{)}^2+(-8-7{)}^2}=\sqrt{1296+225}=\sqrt{1521}=39,andc=AB=\sqrt{(20+36{)}^2+(7-7{)}^2}=\sqrt{(56{)}^2}=56,\text{The coordinates of the centre of the triangle are}(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c})or,\left[\begin{array}{c}\frac{25\times (-36)+39\times 20+56\times 0}{25+39+56},\frac{25\times 7+39\times 7+56\times (-8)}{25+39+56}\end{array}\right]or,\left[\begin{array}{c}\frac{-990+780}{120},\frac{175+273-448}{120}\end{array}\right]or,\left[\begin{array}{c}\frac{-120}{120},\frac{0}{120}\end{array}\right]or,(-1,0)\text{Hence,the coordinates of the centre of the circle inscribed in a triangle whose vertices are}(-36,7),(20,7)and(0,8)is(-1,0).$