Let A(–36, 7), B(20, 7) and C(0, –8) be the vertices of the given triangle.Then, a = BC = \(\sqrt{(0-20)^2+(-8-7)^2}\) = \(\sqrt{400+225}\) = \(\sqrt{625}\) = 25b = AC =\(\sqrt{(0-36)^2+(-8-7)^2}\) = \(\sqrt{1296+225}\) = \(\sqrt{1521}\) = 39c = AB = \(\sqrt{(20+36)^2+(7-7)^2}\) = \(\sqrt{56^2}\) = 56.The co-ordinates of the incentre of the ΔABC are\(\bigg[rac{ax_1+bx_2+cx_3}{a+b+c},rac{ay_1+by_2+cy_3}{a+b+c}\bigg]\) = \(\bigg(rac{25(-36)+39(20)+56 imes0}{25+39+56},rac{25(7)+39(7)+56(-8)}{25+39+56}\bigg)\)= \(\bigg(rac{-900+780}{120},rac{175+273-448}{120}\bigg)\) = \(\bigg(rac{-120}{120},rac{0}{120}\bigg)\) = (–1, 0)