Thank you for reporting, we will resolve it shortly
Q.
Two sides of a parallelogram are along the
lines $4 x+5 y=0$ and $7 x+2 y=0$. If the equation of oneof the diagonals of the parallelogram is $11 x+7 y=9$, then other diagonal passes throughthe point :
Both the lines pass through origin
point $D$ is equal of intersection of $4 x+5 y=0 \,\& \,11 x+7 y=9$
So, coordinates of point $D=\left(\frac{5}{3},-\frac{4}{3}\right)$
Also, point $B$ is point of intersection of $7 x+2 y=0 \,\&\, 11 x+7 y=9$
So, coordinates of point $B=\left(-\frac{2}{3}, \frac{7}{3}\right)$
diagonals of parallelogram intersect at middle let middle point of $B, D$ $\Rightarrow\left(\frac{\frac{5}{3}-\frac{2}{3}}{2}, \frac{-4}{3}+\frac{7}{3}\right)=\left(\frac{1}{2}, \frac{1}{2}\right)$
equation of diagonal $A C$
$\Rightarrow(y-0)=\frac{\frac{1}{\alpha}-0}{\frac{1}{\alpha}-0}(\pi-0)$
$y=x$
diagonal AC passes through $(2,2)$
