The area of a triangle with vertices (a, b+c) | vedclass

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(A) Let the vertices of the triangle be $A(x_1, y_1) = (a, b+c)$,$B(x_2, y_2) = (b, c+a)$,and $C(x_3, y_3) = (c, a+b)$.The area of a triangle with ver

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Let the vertices of the triangle be $A(x_1, y_1) = (a, b+c)$,$B(x_2, y_2) = (b, c+a)$,and $C(x_3, y_3) = (c, a+b)$.The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by the formula:$\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$Substituting the given coordinates into the formula:$\Delta = \frac{1}{2} |a(c+a - (a+b)) + b(a+b - (b+c)) + c(b+c - (c+a))|$Simplify the terms inside the brackets:$\Delta = \frac{1}{2} |a(c-b) + b(a-c) + c(b-a)|$Expand the terms:$\Delta = \frac{1}{2} |ac - ab + ab - bc + bc - ac|$All terms cancel out:$\Delta = \frac{1}{2} |0| = 0$Thus,the area of the triangle is $0$.

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

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