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If the midpoints of the sides AB and AC of triangle ABC are (3,5) and (-3,-3), respectively, then find the length of BC.

\(\text{Let $P_1 =(x_1,\ y_1)=(3,\ 5)$ } \\ \text{Let $P_2 =(x_2,\ y_2)=(-3,\ -3)$ }\)

\(\begin{array}{|rcll|} \hline \dfrac{x_A+x_B}{2} &=& x_1 \\ x_A &=& 2x_1-x_B \\ &=& 2*3-x_B \\ \mathbf{x_A} &=& \mathbf{6-x_B} \\\\ \dfrac{y_A+y_B}{2} &=& y_1 \\ y_A &=& 2y_1-y_B \\ &=& 2*5-x_B \\ \mathbf{y_A} &=& \mathbf{10-y_B} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \dfrac{x_A+x_C}{2} &=& x_2 \\ x_A &=& 2x_2-x_C \\ &=& 2*(-3)-x_C \\ \mathbf{x_A} &=& \mathbf{-6-x_C} \\\\ \dfrac{y_A+y_C}{2} &=& y_2 \\ y_A &=& 2y_2-y_C \\ &=& 2*(-3)-y_C \\ \mathbf{y_A} &=& \mathbf{-6-y_C} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline x_A = 6-x_B &=& -6-x_C \\ \mathbf{x_B -x_C} &=& \mathbf{12} \\\\ y_A = 10-y_B &=& -6-y_C \\ \mathbf{y_B -y_C} &=& \mathbf{16} \\\\ BC &=& \sqrt{12^2+16^2} \\ \mathbf{BC} &=& \mathbf{20} \\ \hline \end{array}\)