Question

find area of triangle with vertices (a, b+c), (b, c+a) and (c, a+b)

Answer 1

hope it may help you

Answer 2

Given:

Vertices of a triangle  are

(a,b+c),(b,c+a) and (c,a+b)

To find:

Area of triangle

Solution:

We know that

Area of triangle is given by

$A=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

Using the formula

Area of triangle is given by

$A=\frac{1}{2}|a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a)|$

$A=\frac{1}{2}|a(c-b)+b(a-c)+c(b-a)|$

$A=\frac{1}{2}|ac-ab+ab-bc+bc-ac|$

$A=\frac{1}{2}(0)=0$

Hence, the area of triangle =0 square units