Question

The lines $ax+by+c=u,bx+cy+a=0$ and $cx+ay+b=0(a\ne b\ne c)$ are concurrent:

• A. $a^3+b^3+c^3+3abc=0$
• B. $a^2+b^2+c^2-3abc=0$
• C. $a+b+c=0$
• D. $noneofthese$

Answer 1

The correct option is C $a+b+c=0$

We have,

$\begin{array}{c}ax+by+c=0\\ bx+xy+a=0\\ x+ay+b=0\\ \left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0\end{array}$

(As lines are con current)

$\begin{array}{c}-\left(a^3+b^3+c^3-3abc\right)=0\\ \frac{\left(a+b+c\right)}{2}\left[{\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2\right]=0\end{array}$

As, $(a\ne b\ne c)$

$\Rightarrow a+b+c=0$

Hence,

Option $C$ is correct answer.