The equations to a pair of opposite sides of parallelogram are $x^2-5x+6=0$ and $y^2-6y+5=0$, the equations to its diagonals are

$x^2-5x+6=0\Rightarrow \left(x-2\right)\left(x-3\right)=0$

$\Rightarrow$two of the parallel sides are x = 2, x = 3

$y^2-6y+5=0\Rightarrow \left(y-1\right)\left(y-5\right)=0$

$\Rightarrow$other two parallel sides are y = 1, y = 5

So, the vertices of the parallelogram are (2, 1), (2, 5), (3, 1), (3, 5). Equation of the diagonal are 

$y-5=\frac{1-5}{3-2}(x-2)$

$\Rightarrow$4x + y = 13  and$y-1=\frac{5-1}{3-2}(x-2)$

$\Rightarrow$y = 4x - 7