Question

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

• A. $a^3+b^3+c^3-3abc=0$
• B. $a^2+b^2+c^2-3abc=0$
• C. $a+b+c=0$
• D. None of these

Answer 1

Answer: (A)

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

Three lines,

$l_{1}x+m_{1}y+n_1=0$

$l_{2}x+m_{2}y+n_2=0$

$l_{3}x+m_{3}y+n_3=0$

are concurrent if

$\Rightarrow \left|\begin{array}{ccc}{l}_1& m_1& n_1\\ l_2& m_2& n_2\\ l_3& m_3& n_3\end{array}\right|=0$

in the question the three lines are

$\Rightarrow ax+by+c=0$

$bx+cy+a=0$

$cx+ay+b=0$

$\therefore$ the lines will be concurrent if

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

Expanding along row 1 , we have

$\Rightarrow a\left|\begin{array}{cc}c& a\\ a& b\end{array}\right|-b\left|\begin{array}{cc}b& a\\ c& b\end{array}\right|+c\left|\begin{array}{cc}b& c\\ c& a\end{array}\right|=0$

$\Rightarrow a(bc-a^2)-b(b^2-ac)+c(ab-c^2)=0$

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

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

$\therefore$ The required condiotion for the three lines to be concurrent is $a^3+b^3+c^3-3abc=0$

Hence, the answer is $a^3+b^3+c^3-3abc=0.$