Question

fibd the area of triangle whose vertices are (a,b+c)(b,c+a)(c,a+b)​

Answer 1

Answer:

    let.... (a,b+c) = (x1,y1)

             (b,c+a) = (x2,y2)

             (c,a+b) = (x3,y3)

we know that......  area = 1/2 (x1(y2-y3)+x2(y3-y1)+x3(y1-y2))

                    area = 1/2 (a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a))

                    area = 1/2 (ac-ab+ab-bc+bc-ac)

                    area = 1/2 * 0

                    area = 0.......ANS