check the picture below.
so, we know that 3.5, 2.5 is the midpoint of each diagonal, keeping in mind that, in a parallelogram, the diagonals bisect each other, namely, they cut each other in two equal halves.
[tex]\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ 2}}\quad ,&{{ 5}})\quad
% (c,d)
C&({{ x}}\quad ,&{{ y}})
\end{array}\qquad
% coordinates of midpoint
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)[/tex]
[tex]\bf \left( \cfrac{x+2}{2}~,~\cfrac{y+5}{2} \right)=\stackrel{midpoint}{(3.5,2.5)}\implies
\begin{cases}
\cfrac{x+2}{2}=3.5\\\\
x+2=7\\
\boxed{x=5}\\
----------\\
\cfrac{y+5}{2}=2.5\\\\
y+5=5\\
\boxed{y=0}
\end{cases}[/tex]
[tex]\bf -------------------------------\\\\
\textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
B&({{ 8}}\quad ,&{{ 8}})\quad
% (c,d)
D&({{ x}}\quad ,&{{ y}})
\end{array}\qquad
% coordinates of midpoint
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)[/tex]
[tex]\bf \left( \cfrac{x+8}{2}~,~\cfrac{y+8}{2} \right)=\stackrel{midpoint}{(3.5,2.5)}\implies
\begin{cases}
\cfrac{x+8}{2}=3.5\\\\
x+8=7\\
\boxed{x=-1}\\
----------\\
\cfrac{y+8}{2}=2.5\\\\
y+8=5\\
\boxed{y=-3}
\end{cases}[/tex]
