Answer :
[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]
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]