[tex]\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ \square }}\quad ,&{{ \square }})\quad
% (c,d)
&({{ \square }}\quad ,&{{ \square }})
\end{array}\qquad
% coordinates of midpoint
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)[/tex]
so.. now, you have the midpoint, and one endpoint, p1 = -3, 6
so... let us use those two fellows
[tex]\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
p1&({{-3 }}\quad ,&{{6}})\quad
% (c,d)
p2&({{ \square }}\quad ,&{{ \square }})
\end{array}\qquad
% coordinates of midpoint
\boxed{(-1,4)}\leftarrow midpoint
\\\\
\textit{that means, that }\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)=(-1,4)
\\\\
or\ that
[/tex]
[tex]\bf \begin{cases}
\cfrac{{{ x_2}} + {{ x_1}}}{2}=-1\implies &\cfrac{\square +(-3)}{2}=1\\
&\uparrow \\
&\textit{solve for }\square \textit{ to get }x_2\\
\\\\
\cfrac{{{ y_2}} + {{ y_1}}}{2} =4\implies &\cfrac{\square +(6)}{2}=4\\
&\uparrow \\
&\textit{solve for }\square \textit{ to get }y_2\\
\end{cases}
\\\\
recall\ that\ p1(x_2,y_2)[/tex]
