if line x+2y-3=0,2x+y-3=0and the line l are concurrent.if the line l passes through origin then its equation

Let concurrent point be (h,k)Then both line will satisfies this point h+2k-3=0 and2h+k-3=0 By solving these two equations we geth=1and k=1 thus , we have two points (0,0) and (1,1) Therefore by two. Point form of lineThe equation of given line would be X-Y-1=0

By using family of lines the equation ofL is (x+2y-3)+∆(2x+y-3)=0 ,the line passes through origin (0,0).We get ∆=-1-x+y=0x=y