The equation of the lines through the point  $\left(2,3\right)$and making an intercept of length 2 units between the lines $y+2x=3\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}y+2x=5$  is

$\text{ As the lines are parallel, distance between them }=\frac{|5-3|}{\sqrt{4+1}}=\frac{2}{\sqrt{3}}=BC$

$\begin{array}{l}\therefore AB=\sqrt{4-\frac{4}{5}}=\frac{4}{\sqrt{5}}\\ \text{ From }\mathrm{△}ABC,\tan \theta =\frac{\left(\frac{2}{\sqrt{5}}\right)}{\left(\frac{4}{\sqrt{5}}\right)}=\frac{1}{2}\end{array}$

Slope of the given line is 2.

$\text{ Let }{m}_1\text{ be the slope of the required line }$

$\begin{array}{l}\text{ Now }\tan \theta =\frac{1}{2}=\left|\frac{m_1+2}{1-2m_1}\right|\\ \Rightarrow \frac{m_1+2}{1-2m_1}=\pm \frac{1}{2}\\ \text{ Consider }\frac{1}{2}=\frac{m_1+2}{1-2m_1}\\ \Rightarrow 1-2m_1=2m_1+4\\ \Rightarrow 4m_1=-3\\ \Rightarrow m_1=-\frac{3}{4}\end{array}$

$\text{ Similarly by taking }-\frac{1}{2}\text{ , other value of }{m}_1\text{ is undefined. }$

$\text{ The equation of the line having slope }m\text{ and Passing through the point }\left(x_1,y_1\right)\text{ is }$

$\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ \Rightarrow y-3=-\frac{3}{4}(x-2)\text{ and }y-3=\frac{1}{0}(x-2)\end{array}$

$\therefore \text{ Equations to the required lines are }3x+4y-18=0\text{ and }x=2$