Given A(3, 4), B(-7, 2), and C(1, -2), find the coordinates of the mid-points K of AB and P of AC, and then find the equation of the perpendicular bisector of the line KP.

The midpoint formula for two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
Given A$(3, 4)$ and B$(-7, 2)$:

$$K = \left(\frac{3 + (-7)}{2}, \frac{4 + 2}{2}\right)$$ $$K = \left(\frac{-4}{2}, \frac{6}{2}\right)$$ $$K = (-2, 3)$$

Given A$(3, 4)$ and C$(1, -2)$:

$$P = \left(\frac{3 + 1}{2}, \frac{4 + (-2)}{2}\right)$$ $$P = \left(\frac{4}{2}, \frac{2}{2}\right)$$ $$P = (2, 1)$$

The coordinates are $\boxed{K(-2, 3) \text{ and } P(2, 1)}$.

Let M be the midpoint of KP. Given K$(-2, 3)$ and P$(2, 1)$:

$$M = \left(\frac{-2 + 2}{2}, \frac{3 + 1}{2}\right)$$ $$M = \left(\frac{0}{2}, \frac{4}{2}\right)$$ $$M = (0, 2)$$

The gradient formula for two points $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

$$m_{KP} = \frac{1 - 3}{2 - (-2)}$$ $$m_{KP} = \frac{-2}{2 + 2}$$ $$m_{KP} = \frac{-2}{4}$$ $$m_{KP} = -\frac{1}{2}$$

The product of the gradients of two perpendicular lines is $-1$.
Let $m_{\perp}$ be the gradient of the perpendicular bisector.

$$m_{KP} \cdot m_{\perp} = -1$$ $$-\frac{1}{2} \cdot m_{\perp} = -1$$ $$m_{\perp} = -1 \cdot (-2)$$ $$m_{\perp} = 2$$

Using the point-slope form $y - y_1 = m(x - x_1)$ with midpoint M$(0, 2)$ and gradient $m_{\perp} = 2$:

$$y - 2 = 2(x - 0)$$ $$y - 2 = 2x$$ $$y = 2x + 2$$

The equation of the perpendicular bisector is $\boxed{y = 2x + 2}$.