Two sides of a parallelogram are along the lines $4x+5y=0$ and $7x+2y=0 .$ If the equation of one of the diagonals of the parallelogram is $11x+7y=9,$ then other diagonal passes through the point:

Both the lines $4x+5y=0,7x+2y=0$ pass through origin.

$D$ is the point of intersection of $4x+5y=0 \& 11x+7y=9$

So upon solving we get the coordinates of point $D=\left(\frac{5}{3},-\frac{4}{3}\right)$

Also, point $B$ is the point of intersection of $7x+2y=0 \& 11x+7y=9$

So, coordinates of point $B=\left(-\frac{2}{3},\frac{7}{3}\right)$

The diagonals of parallelogram bisect each other, so the mid point of $BD$ is

$\left(\frac{\frac{5}{3}-\frac{2}{3}}{2},\frac{\frac{-4}{3}+\frac{7}{3}}{2}\right)=\left(\frac{1}{2},\frac{1}{2}\right)$

The equation of diagonal $AC$ is

$\left(y-0\right)=\frac{\frac{1}{2}-0}{\frac{1}{2}-0}\left(x-0\right)$ i.e., $y=x$

Therefore, the diagonal $AC$ passes through $\left(2,2\right).$