Question

Question 18
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) $abc$

Answer 1

$\text{Vertices of the given triangle are:}A=(x_1,y_1)=(a,b,c)B=(x_2,y_2)=(b,c+a)C=(x_3,y_3)=(c,a+b)Areaof\Delta ABC=\Delta =\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]\therefore \Delta =\frac{1}{2}\left[a\right(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a\left)\right]=\frac{1}{2}\left[a\right(c-b)+b(a-c)+c(b-a\left)\right]=\frac{1}{2}(ac-ab+ab-bc+bc-ac)=\frac{1}{2}\left(0\right)=0\text{Hence, area of the given triangle is 0.}$