Two consecutive sides of a parallelogram are 4x + 5y = 0 and 7x + 2y = 0. If the equation to one diagonal is 11x + 7y = 9, then the equation of the other diagonal is 

$\text{ Let the equation of sides }AB\text{ and }AD\text{ of the parallelogram ABCD be}$

$4x+5y=0----\left(1\right)$

$\text{ and }7x+2y=0---\left(2\right)$

respectively. Solving (l) and (2), we have x = 0, y= 0

 A = (0,0) Equation of one diagonal of the parallelogram is

$11x+7y=9---\left(3\right)$

Clearly, I (0, 0) does not lie on diagonal (3), therefore (3) is the equation of diagonal BD.
$\text{ Solving (1) and (3), we get }B\equiv \left(\frac{5}{3},-\frac{4}{3}\right)\text{ and }D\equiv \left(-\frac{2}{3},\frac{7}{3}\right)\text{ . }$

$\text{ Since }H\text{ is the middle point of }BD\therefore H\equiv \left(\frac{1}{2},\frac{1}{2}\right)\text{ . }$

Now the equation of diagonal AC which passes through 

$A(0,0)\text{ and }H\left(\frac{1}{2},\frac{1}{2}\right)\text{ is }y-0=\frac{0-\frac{1}{2}}{0-\frac{1}{2}}(x-0)$

$\text{ or }y-x=0$