Question

The area of a triangle with vertices $(a,b+c),(b,c+a)$ and $(c,a+b)$ is

• A. $a$
• B. $b$
• C. $c$
• D. $0$

Answer 1

Answer: (D)

The area of $△ABC$ with vertices $A\equiv (x_1,y_1)$, $B\equiv (x_2,y_2)$ and $C\equiv (x_3,y_3)$ is given as,

$A\left(△ABC\right)=\left|\frac{1}{2}\right[x_1(y_3-y_2)+x_2(y_1-y_3)+x_3(y_2-y_1)\left]\right|$

In this problem,

$A\left(△ABC\right)=\left|\frac{1}{2}\right[a(a+b-(c+a\left)\right)+b(b+c-(a+b\left)\right)+c(c+a-(b+c\left)\right)\left]\right|$

$\therefore A\left(△ABC\right)=\left|\frac{1}{2}\right[ab-ac+bc-ba+ca-cb\left]\right|$

$\therefore A\left(△ABC\right)=0$

Hence, the correct Option is B.