Question

Find the area of the triangle whose vertices are (a,b+c),(b,c+a) and (c,a+b).

Answer 1

concept : if three points A$(x_1,y_1)$ , B$(x_2,y_2)$ and C$(x_3,y_3)$ form a triangle ABC.

then, area of triangle ABC = $\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$

here given, three points (a, b + c), (b, c + a) and (c, a + b)

then, area of triangle formed by these points = 1/2 [ a(c + a - a - b) + b(a + b - b - c) + c(b + c - c - a)]

= 1/2 [a(c - b) + b(a - c) + c(b - a)]

= 1/2 [ ac - ab + ba - bc + cb - ac ]

= 0

hence, area of triangle whose vertices are (a, b + c), (b, c + a) and (c, a + b) = 0

[important point : if area of triangle becomes zero, it means vertices of triangle is collinear. I mean, they lie in same line . ]

Answer 2

Answer:

here is ur answer.........