A line T passes through (5, 4) and (1, 2). a) Find the equation of T in the form y = mx + c. b) Find the equation of line D, the perpendicular bisector of line T in the form ax + by + c = 0.

$$y - 2 = \frac{1}{2}(x - 1)$$ $$y - 2 = \frac{1}{2}x - \frac{1}{2}$$

$$y = \frac{1}{2}x - \frac{1}{2} + 2$$ $$y = \frac{1}{2}x + \frac{4}{2} - \frac{1}{2}$$ $$y = \frac{1}{2}x + \frac{3}{2}$$

The equation of line T is $\boxed{y = \frac{1}{2}x + \frac{3}{2}}$.

The perpendicular bisector passes through the midpoint of the segment connecting $(5, 4)$ and $(1, 2)$.

$$\text{Midpoint} = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) = \left(\frac{5+1}{2}, \frac{4+2}{2}\right) = \left(\frac{6}{2}, \frac{6}{2}\right) = (3, 3)$$

Line D is perpendicular to line T. The slope of T is $m_T = \frac{1}{2}$.
The slope of a perpendicular line $m_D$ is the negative reciprocal of $m_T$.

$$m_D = -\frac{1}{m_T} = -\frac{1}{1/2} = -2$$

$$y - 3 = -2(x - 3)$$ $$y - 3 = -2x + 6$$

$$2x + y - 3 - 6 = 0$$ $$2x + y - 9 = 0$$

The equation of line D is $\boxed{2x + y - 9 = 0}$.

Line Y is parallel to line T, so its slope $m_Y$ is equal to $m_T$.

$$m_Y = m_T = \frac{1}{2}$$

Line Y passes through the point $(-8, 0)$. Using the point-slope form:

$$y - 0 = \frac{1}{2}(x - (-8))$$ $$y = \frac{1}{2}(x + 8)$$ $$y = \frac{1}{2}x + 4$$

Equation of line Y: $y = \frac{1}{2}x + 4$
Equation of line D (from part b, rearranged to $y=mx+c$ form): $y = -2x + 9$

Set the expressions for $y$ equal to each other:

$$\frac{1}{2}x + 4 = -2x + 9$$

Multiply the entire equation by 2 to eliminate the fraction.

$$2\left(\frac{1}{2}x + 4\right) = 2(-2x + 9)$$ $$x + 8 = -4x + 18$$ $$x + 4x = 18 - 8$$ $$5x = 10$$ $$x = \frac{10}{5}$$ $$x = 2$$

$$y = \frac{1}{2}(2) + 4$$ $$y = 1 + 4$$ $$y = 5$$

The coordinates of Z are $\boxed{(2, 5)}$.