28)
[tex]\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ 1}}\quad ,&{{ 7}})\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)
\\\\\\
\left( \cfrac{x+1}{2}~,~\cfrac{y+7}{2} \right)=\stackrel{B}{(-3,1)}\implies
\begin{cases}
\cfrac{x+1}{2}=-3\\\\
x+1=-6\\
\boxed{x=-7}\\
----------\\
\cfrac{y+7}{2}=1\\\\
y+7=2\\
\boxed{y=-5}
\end{cases}[/tex]
29)
[tex]\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ x}}\quad ,&{{ y}})\quad
% (c,d)
C&({{ -5}}\quad ,&{{ 4}})
\end{array}\qquad
% coordinates of midpoint
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left( \cfrac{-5+x}{2}~,~\cfrac{4+y}{2} \right)=\stackrel{B}{(-2,5)}\implies
\begin{cases}
\cfrac{-5+x}{2}=-2\\\\
-5+x=-4\\
\boxed{x=1}\\
-------\\
\cfrac{4+y}{2}=5\\\\
4+y=10\\
\boxed{y=6}
\end{cases}[/tex]
so.. let's also do 32)
32)
[tex]\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ x}}\quad ,&{{ y}})\quad
% (c,d)
C&({{ \frac{5}{3}}}\quad ,&{{ -6}})
\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{\frac{5}{3}+x}{2}~,~\cfrac{-6+y}{2} \right)=\stackrel{B}{\left( \frac{8}{3},4 \right)}\implies
\begin{cases}
\cfrac{\frac{5}{3}+x}{2}=\cfrac{8}{3}\\\\
\frac{5}{3}+x=\frac{16}{3}\\
x=\frac{16}{3}-\frac{5}{3}\\\\
\boxed{x=\frac{11}{3}}\\
------\\
\cfrac{-6+y}{2}=4\\\\
-6+y=8\\
\boxed{y=14}
\end{cases}[/tex]
