Question

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

• A. 0
• B. 1
• C. a + b + c
• D. a² + b² + c²

Answer 1

The correct option is A

0

Area of the triangle = $\frac{1}{2}$[$x_1$($y_2$ - $y_3$) + $x_2$ ($y_3$ - $y_1$)+ $x_3$($y_1$ - $y_2$)]

So, $\frac{1}{2}$[a(c+a-a-b) + b(a+b-b-c) + c(b+c-c-a)

= $\frac{1}{2}$[ ac - ab + ab - bc + bc - ac]

= 0