Question

The three distinct straight lines $ax+by+c=0$;$bx+cy+a=0$ and $cx+ay+b=0$ are concurrent then

• A. $a+b+c=0$
• B. $a^3+b^3+c^3=3abc$
• C. $a=b=c$
• D. $a^2+b^2+c^2=ab+bc+ca$

Answer 1

Answer: (A), (B)

The correct options are
A $a+b+c=0$
B $a^3+b^3+c^3=3abc$

The lines are concurrent,
$=>\left[\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right]=0$
Applying,$C_1=C_1+C_2+C_3$,
$=\left[\begin{array}{ccc}a+b+c& b& c\\ a+b+c& c& a\\ a+b+c& a& b\end{array}\right]$
Applying, $R_1=R_1-R_2$ and $R_2=R_2-R_3$,
$=(a+b+c)\left[\begin{array}{ccc}0& b-c& c-a\\ 0& c-a& a-b\\ 1& a& b\end{array}\right]$
Expanding by $C_1$,
$=(a+b+c)(-a^2-b^2-c^2+ab+bc+ac)=0$
$=>(a+b+c)=0$
or,
$(a+b+c)(-a^2-b^2-c^2+ab+bc+ac)=0$
$abc-a^3-b^3+abc+abc-c^3=0$
$a^3+b^3+c^3=3abc$
So options are A and B.