ABCD is a parallelogram. Equations of $\overleftrightarrow{AB}and\overleftrightarrow{AD}$  are $4x+5y=0,7x+2y=0$  and the equation of the diagonal $\overleftrightarrow{BD}$  is $11x+7y=9$ . Then the equation of  $\overleftrightarrow{AC}$ is

$\text{Equation of}\overline{AB}\text{is}4x+5y=0\dots \text{(1)}$
$\text{Equation of}\overline{AD}\text{is}7x+2y=0\dots \text{(2)}$
$\text{Equation of}\overline{BD}\text{is}11x+7y=9\dots \text{(3)}$
$\text{Point of intersection of (1) and (2) is}A(0,0)$
$\text{Point of intersection of (1) and (3) is}B=(\frac{5}{3},\frac{-4}{3})$
$\text{Point of intersection of (2) and (3) is}D=(\frac{-2}{3},\frac{7}{3})$
$\text{Now,}C=B+D-A=(1,1)$
$\text{Equation of}\overline{AC}\text{is}x=y$