jmc

The midpoint $M$ of line segment $\overline{BC}$ is $\left(\frac{35}{2},\frac{39}{2}\right)$. The equation of the median can be found by $-5=\frac{q-\frac{39}{2}}{p-\frac{35}{2}}$. Cross multiply and simplify to yield that $-5p+\frac{35\cdot 5}{2}=q-\frac{39}{2}$, so $q=-5p+107$. Use determinants to find that the area of $△ABC$ is $\frac{1}{2}$ (note that there is a missing absolute value; we will assume that the other solution for the triangle will give a smaller value of $p+q$, which is provable by following these steps over again). We can calculate this determinant to become $140=$ $⟹140=240-437-20p+23q+19p-12q$ $=-197-p+11q$. Thus, $q=\frac{1}{11}p-\frac{337}{11}$. Setting this equation equal to the equation of the median, we get that $\frac{1}{11}p-\frac{337}{11}=-5p+107$, so $\frac{56}{11}p=\frac{107\cdot 11+337}{11}$. Solving produces that $p=15$. Substituting backwards yields that $q=32$; the solution is $p+q=\boxed{47}$.