Two sides of a parallelogram are along the lines 4x + 5y = 0 and 7x...

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

$D$ is the point of intersection of $4x + 5y = 0 \text{ and } 11x + 7y = 9$

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

Also, point $B$ is the point of intersection of $7x + 2y = 0 \text{ and } 11x + 7y = 9$

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

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

\left(\dfrac{\dfrac{5}{3} - \dfrac{2}{3}}{2}, \dfrac{\dfrac{- 4}{3} + \dfrac{7}{3}}{2}\right) = \left(\dfrac{1}{2}, \dfrac{1}{2}\right)$</p><p>The equation of diagonal$AC$is</p><p>\left(y - 0\right) = \dfrac{\dfrac{1}{2} - 0}{\dfrac{1}{2} - 0}\left(x - 0\right)$ i.e., $y = x$

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