[Triangle](https://artofproblemsolving.com/wiki/index.php/Triangle) $ABC$ lies in the [cartesian plane](https://artofproblemsolving.com/wiki/index.php/Cartesian_plane) and has an [area](https://artofproblemsolving.com/wiki/index.php/Area) of $70$. The coordinates of $B$ and $C$ are $(12,19)$ and $(23,20),$ respectively, and the coordinates of $A$ are $(p,q).$ The [line](https://artofproblemsolving.com/wiki/index.php/Line) containing the [ median](https://artofproblemsolving.com/wiki/index.php/Median_of_a_triangle) to side $BC$ has [slope](https://artofproblemsolving.com/wiki/index.php/Slope) $-5.$ Find the largest possible value of $p+q.$
The [midpoint](https://artofproblemsolving.com/wiki/index.php/Midpoint) $M$ of [line segment](https://artofproblemsolving.com/wiki/index.php/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](https://artofproblemsolving.com/wiki/index.php/Determinant) to find that the [area](https://artofproblemsolving.com/wiki/index.php/Area) of $\triangle ABC$ is $\frac{1}{2} \begin{vmatrix}p & 12 & 23 \\  q & 19 & 20 \\ 1 & 1 & 1\end{vmatrix} = 70$ (note that there is a missing [absolute value](https://artofproblemsolving.com/wiki/index.php/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) (alternatively, we could use the [Shoelace Theorem](https://artofproblemsolving.com/wiki/index.php/Shoelace_Theorem)). We can calculate this determinant to become $140 = \begin{vmatrix} 12 & 23 \\ 19 & 20 \end{vmatrix} - \begin{vmatrix} p & q \\ 23 & 20 \end{vmatrix} + \begin{vmatrix} p & q \\ 12 & 19 \end{vmatrix}$ $\Longrightarrow 140 = 240 - 437 - 20p + 23q + 19p - 12q$ $= -197 - p + 11q$. 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](https://artofproblemsolving.com/wiki/index.php/Substitution) backwards yields that $q = 32$; the solution is $p + q = \boxed{047}$.