Find the coordinates of C if B is the Midpoint of AC.
B(1,6)
A(3,-2)
Midpoint of AB=(2,2)
To find the midpoint between 2 points, we can do $${\frac{\left({{\mathtt{x}}}^{{\mathtt{1}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{x}}}^{{\mathtt{2}}}\right)}{{\mathtt{2}}}}$$,$${\frac{\left({{\mathtt{y}}}^{{\mathtt{1}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{y}}}^{{\mathtt{2}}}\right)}{{\mathtt{2}}}}$$.
So by inputting the coordinates you gave, $${\frac{\left({\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}\right)}{{\mathtt{2}}}} = {\mathtt{1}}$$,$${\frac{\left({\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{y}}\right)}{{\mathtt{2}}}} = {\mathtt{6}}$$.
Then we can simply solve for x and y.
$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}={\mathtt{2}}\\
{\mathtt{x}}=-{\mathtt{1}}\end{array}\right)} \Rightarrow {\mathtt{x}} = -{\mathtt{1}}$$
$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{y}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{y}}={\mathtt{12}}\\
{\mathtt{y}}={\mathtt{14}}\end{array}\right)} \Rightarrow {\mathtt{y}} = {\mathtt{14}}$$
So C= (-1,14)
To find the midpoint between 2 points, we can do $${\frac{\left({{\mathtt{x}}}^{{\mathtt{1}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{x}}}^{{\mathtt{2}}}\right)}{{\mathtt{2}}}}$$,$${\frac{\left({{\mathtt{y}}}^{{\mathtt{1}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{y}}}^{{\mathtt{2}}}\right)}{{\mathtt{2}}}}$$.
So by inputting the coordinates you gave, $${\frac{\left({\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}\right)}{{\mathtt{2}}}} = {\mathtt{1}}$$,$${\frac{\left({\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{y}}\right)}{{\mathtt{2}}}} = {\mathtt{6}}$$.
Then we can simply solve for x and y.
$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}={\mathtt{2}}\\
{\mathtt{x}}=-{\mathtt{1}}\end{array}\right)} \Rightarrow {\mathtt{x}} = -{\mathtt{1}}$$
$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{y}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{y}}={\mathtt{12}}\\
{\mathtt{y}}={\mathtt{14}}\end{array}\right)} \Rightarrow {\mathtt{y}} = {\mathtt{14}}$$
So C= (-1,14)