Question

Consider the points $A(a,b+c)$, $B(b,c+a)$, and $C(c,a+b)$ be the vertices of $△$ABC. The area of $△$ABC is:

• A. $2(a^2+b^2+c^2)$
• B. $a^2+b^2+c^2$
• C. $2(ab+bc+ca)$
• D. None of these

Answer 1

Answer: (D)

None of these

We know that the area of the triangle whose vertices are $(x_1,y_1),(x_2,y_2),$ and $(x_3,y_3)$ is $\frac{1}{2}\left|x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right|$

Since, vertices are $A(a,b+c),B(b,c+a),$ and $C(c,a+b)$
$\therefore$ Area $△ABC=\frac{1}{2}|a(c+a)-b(b+c)+b(a+b)-c(c+a)+c(b+c)-a(a+b)|=0$ .

Hence option 'D' is correct.