If the equations of the pairs of opposite sides of a parallelogram...

Equations of the sides of the parallelogram are $(x-3)(x-2)=0$ and $(y-5)(y-1)=0$ i.e. $x=3, x=2 ; y=5, y=1$ Hence its vertices are : ($\mathrm{A}($2,1) ; $\mathrm{B}($3,1)); ($\mathrm{C}($3,5) ; $\mathrm{D}($2,5)) Equation of the diagonal ($\mathrm{AC}$) is (y-1=$\frac{4}{1}($x-2) \Rightarrow y=4 x-7) Equation of the diagonal ($\mathrm{BD}$) is (y-1=$\frac{4}{-1}($x-3) \Rightarrow 4 x+y=13)