1. **State the problem:** We have points $P(-5,a)$ and $Q(7,3a)$ on a coordinate plane, with $a>0$. The length of segment $PQ$ is given as $4\sqrt{10}$. We need to find the equation of the perpendicular bisector of $PQ$ in the form $y=mx+c$ where $m$ and $c$ are integers. 2. **Use the distance formula to find $a$:** The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$ Here, $$PQ=4\sqrt{10}$$ $$\Rightarrow 4\sqrt{10} = \sqrt{(7 - (-5))^2 + (3a - a)^2}$$ $$= \sqrt{(7+5)^2 + (2a)^2} = \sqrt{12^2 + 4a^2} = \sqrt{144 + 4a^2}$$ 3. **Square both sides to eliminate the square root:** $$ (4\sqrt{10})^2 = 144 + 4a^2 $$ $$ 16 \times 10 = 144 + 4a^2 $$ $$ 160 = 144 + 4a^2 $$ 4. **Solve for $a^2$:** $$ 160 - 144 = 4a^2 $$ $$ 16 = 4a^2 $$ $$ \cancel{4} \times 4 = \cancel{4} a^2 $$ $$ 4 = a^2 $$ 5. **Find $a$ (since $a>0$):** $$ a = 2 $$ 6. **Find coordinates of $P$ and $Q$ with $a=2$:** $$ P(-5, 2), \quad Q(7, 6) $$ 7. **Find midpoint $M$ of $PQ$:** Midpoint formula: $$ M = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) $$ $$ M = \left( \frac{-5+7}{2}, \frac{2+6}{2} \right) = (1, 4) $$ 8. **Find slope of $PQ$:** Slope formula: $$ m_{PQ} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - 2}{7 - (-5)} = \frac{4}{12} = \frac{1}{3} $$ 9. **Find slope of perpendicular bisector:** The perpendicular slope is the negative reciprocal: $$ m = -\frac{1}{m_{PQ}} = -3 $$ 10. **Write equation of perpendicular bisector passing through midpoint $M(1,4)$:** Using point-slope form: $$ y - y_1 = m(x - x_1) $$ $$ y - 4 = -3(x - 1) $$ $$ y - 4 = -3x + 3 $$ 11. **Simplify to slope-intercept form $y=mx+c$:** $$ y = -3x + 3 + 4 $$ $$ y = -3x + 7 $$ **Final answer:** $$ \boxed{y = -3x + 7} $$