If 3a+2b+4c=0 then show that ax+by+c=0 represents a family of concurrent lines and find the point of concurrency.

3a + 2b + 4c = 0 $\Rightarrow$3a + 2b = –4c
$$\begin{gathered} \begin{array}{l}\Rightarrow c=-\frac{(3\mathrm{a}+2\mathrm{b})}{4}\\ \mathrm{ax}+\mathrm{by}+\mathrm{c}=0\Rightarrow \mathrm{ax}+\mathrm{by}-\left(\frac{3\mathrm{a}+2\mathrm{b}}{4}\right)=0\end{array} \\ \mathrm{a}\left(\mathrm{x}-\frac{3}{4}\right)+\mathrm{b}\left(\mathrm{y}-\frac{1}{2}\right)=0 \end{gathered}$$
$\therefore \mathrm{ax}+\mathrm{by}+\mathrm{c}=0 \mathrm{represents} \mathrm{a} \mathrm{set} \mathrm{of} \mathrm{lines} \mathrm{passing} \mathrm{through} \mathrm{the} \mathrm{point} \mathrm{of} \mathrm{intersection} \mathrm{of}$
$$\begin{gathered} \mathrm{x}-\frac{3}{4}=0,\mathrm{y}-\frac{1}{2}=0\Rightarrow \mathrm{x}=\frac{3}{4},\mathrm{y}=\frac{1}{2} \\ \text{ ∴point of concurrence }=\left(\frac{3}{4},\frac{1}{2}\right) \end{gathered}$$