#### The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is(A) (a+b+c)2(B) 0(C)( a + b + c)(D) abc

Vertices are (a, b + c), (b, c + a), (c, a + b)

Here $A(x\textsubscript{1}, y\textsubscript{1}) = (a , b + c)\\$

$\\B\left( {{x}_{2}},{{y}_{2}} \right)= (b , c + a)\\ C\left( {{x}_{3}},{{y}_{3}} \right)= (c , a + b)\\$

We know that

$Area of triangle = \frac{1}{2} [x\textsubscript{1}(y\textsubscript{2} - y\textsubscript{3}) + x\textsubscript{2}(y\textsubscript{3} - y\textsubscript{1}) + x\textsubscript{3}(y\textsubscript{1} - y\textsubscript{2})]\\$
$\\ = \frac{1}{2} [a(c + a - a - b) + b(a + b - b - c) + c(b + c - c -a)]\\ = \frac{1}{2} [a(c - b) + b(a - c) + c(b - a)]\\ = \frac{1}{2} [ac - ab + ab - bc + bc - ac] = 0\\$

Hence, option B is correct.