Transofrm the equation $(2+5\mathrm{k})\mathrm{x}-3(1+2\mathrm{k})\mathrm{y}+(2-\mathrm{k})=0\text{ into the form of }{\mathrm{L}}_1+{\mathrm{\lambda L}}_2=0\text{ and }$find the point of concurrency of the family of straight lines.

$\begin{array}{l}(2+5\mathrm{k})\mathrm{x}-3(1+2\mathrm{k})\mathrm{y}+(2-\mathrm{k})=0\\ 2\mathrm{x}-3\mathrm{y}+2+\mathrm{k}(5\mathrm{x}-6\mathrm{y}-1)=0\dots \left(1\right)\end{array}$
$\text{ eq'n (1) Transform into }{\mathrm{L}}_1+{\mathrm{\lambda L}}_2=0$
Where  ${\mathrm{L}}_1=2\mathrm{x}-3\mathrm{y}+2=0;{\mathrm{L}}_2=5\mathrm{x}-6\mathrm{y}-1=0$
$$\begin{gathered} \text{By Solving }{\mathrm{L}}_1\mathrm{\&}{\mathrm{L}}_2\mathrm{we} \mathrm{get} \\ (\mathrm{x},\mathrm{y})=(5,4) \\ \therefore point of intersection=(5,4) \end{gathered}$$