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  • 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

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

Answers (1)

Vertices are (a, b + c), (b, c + a), (c, a + b)

Here A(x\textsubscript{1}, y\textsubscript{1}) = (a , b + c)\\

\\B\left( {{x}_{2}},{{y}_{2}} \right)= (b , c + a)\\ C\left( {{x}_{3}},{{y}_{3}} \right)= (c , a + b)\\

We know that 

$Area of triangle $= \frac{1}{2} [x\textsubscript{1}(y\textsubscript{2} - y\textsubscript{3}) + x\textsubscript{2}(y\textsubscript{3} - y\textsubscript{1}) + x\textsubscript{3}(y\textsubscript{1} - y\textsubscript{2})]\\
\\ = \frac{1}{2} [a(c + a - a - b) + b(a + b - b - c) + c(b + c - c -a)]\\ = \frac{1}{2} [a(c - b) + b(a - c) + c(b - a)]\\ = \frac{1}{2} [ac - ab + ab - bc + bc - ac] = 0\\

Hence, option B is correct.

Posted by

infoexpert21

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