Question

If $a+b+c=0$and $p\ne 0$ the lines $ax+(b+c)y=p$, $bx+(c+a)y=p$ and $cx+(a+b)y=p$

• A. $Donotintersect$
• B. $Intersect$
• C. $Areconcurrent$
• D. $Noneofthese$

Answer 1

The correct option is C

$Areconcurrent$

Explanation for correct option :

Given, $a+b+c=0$

Equation of line is $ax+(b+c)y=p$

$\begin{array}{rcl}ax-ay& =& p....\left(i\right)\because a+b+c=0\\ bx+(c+a)y& =& p\\ bc-by& =& p....\left(ii\right)\\ cx+(a+b)y& =& p\\ cx–cy& =& p.\dots \left(iii\right)\\ ∆& =& \left[\begin{array}{ccc}a& -a& p\\ b& -b& p\\ c& -c& p\end{array}\right]\\ & =& -p\left[\begin{array}{ccc}a& a& 1\\ b& b& 1\\ c& c& 1\end{array}\right]\\ & =& 0\\ & & \end{array}$

The line are concurrent.

Hence, the correct option is $\left(C\right)$