The set of lines $a x+b y+c=0$ where $3 a+2 b+4 c=0$ is concurrent at the point

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Professor PC Thomas Classes · Answer

$$ \begin{aligned} & 	ext \quad c=\frac{-3 a-2 b}{4} \& 	herefore \quad a x+b y+c=0 \& \Rightarrow \quad a x+b y+\left(\frac{-3 a-2 b}{4}ight)=0 \& \Rightarrow \quad a(4 -3)+b(4 y-2)=0\end{aligned}$$This line passes through $\left(\frac{3}{4}, \frac{1}{2}ight)$ for all values of $a$ and $b$.\ $	herefore \quad$ Point of concurrence of the lines is $\left(\frac{3}{4}, \frac{1}{2}ight)$.\