Question

If the lines $ax+by+c=0,bx+cy+a=0$ and $cx+ay+b=0$ be concurrent, then

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

Answer 1

Answer: (C)

$a^3+b^3+c^3-3abc=0$

For lines to be concurrent,

$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$

On solving, we get

$a^3+b^3+c^3-3abc=0$