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

Question

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

Infinity Learn · Accepted Answer

Let ∆be the area of the triangle with vertices at  (a$$, b $$+$$ $$c), (b$$, c $$+$$ $$a) and (c$$, $$a+$$ $$b), Then,∆$$=$$ absolute value of $$12a$$    $$b+c$$    $$1b$$    $$c+a$$    $$1c$$    $$a+b$$    $$1$$⇒∆$$=$$ Absolute value of $$f12a$$    $$a+b+c$$    $$1b$$    $$b+c+a$$    $$1c$$    $$c+a+b$$    $$1$$ Applying $$C2$$→$$C2+C1$$⇒∆$$=$$ Absolute value of $$12(a+b+c)a$$    $$1$$    $$1b$$    $$1$$    $$1c$$    $$1$$    $$1$$                                                                       [Taking $$a+b+c$$ common from $$C2$$ ] ⇒ Δ$$=0$$

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

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

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