Question

The straight lines $4ax+3by+c=0,$where $a+b+c=0,$ are concurrent, at the point

• A. $(4,3)$
• B. $\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.,\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)$
• C. $\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)$
• D. None of these

Answer 1

The correct option is B

$\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.,\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)$

Explanation for the correct option:

Given: $4ax+3by+c=0...\left(i\right)$

$a+b+c=0...\left(ii\right)$

$\Rightarrow c=-(a+b)$

Put $c$in $\left(i\right)$

$$\begin{gathered} \Rightarrow 4ax+3by–a–b=0 \\ \Rightarrow a(4x-1)+b(3y-1)=0 \\ \Rightarrow (4x-1)+\left(\raisebox{1ex}{$b$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\right)\left(3y-1\right)=0 \\ \Rightarrow (4x-1)=0,(3y-1)=0 \\ \Rightarrow x=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.,y=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right. \end{gathered}$$

Hence, the correct option is option B