Question

The area of ∆ABC with vertices A(a, b + c), B(b, c + a) and C(c, a + b) is

(a) (a + b + c)²
(b) a + b + c
(c) abc
(d) 0

Answer 1

(d) 0
The given points are $A\left(a,b+c\right),B\left(b,c+a\right)\text{and}C\left(c,a+b\right)$.
Here, $\left(x_1=a,y_1=b+c\right),\left(x_2=b,y_2=c+a\right)\text{and}\left(x_3=c,y_3=a+b\right)$
Therefore,
$\text{Area of}∆ABC=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$

$$\begin{gathered} =\frac{1}{2}\left[a\left\{\left(c+a\right)-\left(a+b\right)\right\}+b\left\{\left(a+b\right)-\left(b+c\right)\right\}+c\left\{\left(b+c\right)-\left(c+a\right)\right\}\right] \\ =\frac{1}{2}\left[a\left(c-b\right)+b\left(a-c\right)+c\left(b-a\right)\right] \\ =\frac{1}{2}\left(ac-ab+ab-bc+bc-ac\right) \\ =0 \end{gathered}$$