Let the equation of two adjacent sides of a parallelogram ABCD be $2\mathrm{x}-3\mathrm{y}=-23$ and $5\mathrm{x}+4\mathrm{y}=23$. If the equation of its one diagonal AC is $3\mathrm{x}+7\mathrm{y}=23$ and the distance of A from the other diagonal is d, then 50d² is equal to _____

$\text{ point of intersection of two given sides is }(-1,7)\text{ which does not lie of AC. }$

A is point of intersection of $2\mathrm{x}-3\mathrm{y}=-23$ and $3\mathrm{x}+7\mathrm{y}=23$

C is point of intersection of $5\mathrm{x}+4\mathrm{y}=23$ and $3\mathrm{x}+7\mathrm{y}=23$

Solve for A & C
$$\begin{gathered} \mathrm{A}(-4,5)\mathrm{C}(3,2) \\ \text{ Mid point }\mathrm{M}\left(-\frac{1}{2},\frac{7}{2}\right) \\ \mathrm{Slope} \mathrm{of} \mathrm{BM}=-7 \\ \mathrm{Equation} \mathrm{of} \mathrm{diagonal} \mathrm{BD} \mathrm{is} \\ \mathrm{BM} \mathrm{y}-7=-7(\mathrm{x}+1) \\ \begin{array}{l}7\mathrm{x}+\mathrm{y}=0\\ \text{ Distance of }\mathrm{A}\text{ from diagonal }\mathrm{BD}\\ \mathrm{d}=\frac{|-28+5|}{\sqrt{50}}=\frac{23}{\sqrt{50}}\\ \therefore 50{\mathrm{d}}^2=529\end{array} \end{gathered}$$