Rewriting the equations we get
[tex]y=-\frac{3}{2}x+4\Rightarrow2y+3x=8,y=-\frac{3}{2}x-1\Rightarrow2y+3x-2[/tex]We know that distance between two parallel lines of the form
[tex]ax+by=c_1,ax+by=c_2[/tex]is
[tex]d=\frac{c_1-c_2}{\sqrt[]{a^2+b^2}}[/tex]Here
[tex]a=3,b=2,c_1=8,c_2=-2[/tex]Thus the distance d will be
[tex]d=\frac{8-(-2)}{\sqrt[]{9+4}}=\frac{10}{\sqrt[]{13}}[/tex]