The area of the triangle with vertices at (a, b + c), (b, c + a) and (c, a+ b), is
0
a+b+c
ab+bc+ca
none of these
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