A line can be represented as
#l-> << p - p_0, vec v >> = 0# which stated that the points along #p-p_0# are orthogonal to #vec v#
For the line #ax+by+c=0# we have
#p_0# is any point obeying the line equation.
#a x_0+b y_0+c=0#
#vec v = (a,b)# and
#p = (x, y)#
so
#l_1->x+2y-5=0#
taking #p_1 = (5,0)# and #vec v_1=(1,2)# we have
#l_1-> << p - p_1, vec v_1 >> = 0#
#l_2->2x+4y-7=0#
taking #p_2=(7/2,0)# and #vec v_2=(2,4)=2 vec v_1# we have
#l_2-> << p-p_2,vec v_2 >> = << p-p_2,2 vec v_1 >> = 0#
so #l_1# and #l_2# are parallel because #vec v_2 = lambda vec v_1# with #lambda ne 0#
being parallels, their distance is easily obtained, computing the projection of #p_2-p_1# into the unit vector #hat(v_1)#. So
#hat(v_1)=(vec v_1)/(norm (vec v_1)) = ((1,2))//sqrt(1+2^2) = (1,2)//sqrt(5)#
#p_2-p_1=(7/2-5,0-0) = (-3/2,0)# and finally
#d = abs(<< p_2-p_1, hat(v_1) >>) = 3/(2sqrt(5))#
