Obtain the equation of the diagonal PR in the form y = mx + c.

The points are $A(-10, -1)$ and $B(2, 7)$.
The midpoint $M$ is given by the formula $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.

$$M = \left(\frac{-10+2}{2}, \frac{-1+7}{2}\right) = \left(\frac{-8}{2}, \frac{6}{2}\right) = (-4, 3)$$

The gradient $m_{AB}$ is given by the formula $m = \frac{y_2-y_1}{x_2-x_1}$.

$$m_{AB} = \frac{7 - (-1)}{2 - (-10)} = \frac{7+1}{2+10} = \frac{8}{12} = \frac{2}{3}$$

Since PR is the perpendicular bisector of AB, its gradient $m_{PR}$ is the negative reciprocal of $m_{AB}$.

$$m_{PR} = -\frac{1}{m_{AB}} = -\frac{1}{\frac{2}{3}} = -\frac{3}{2}$$

The line PR passes through the midpoint $M(-4, 3)$ and has a gradient of $-\frac{3}{2}$. Using the point-slope form $y - y_1 = m(x - x_1)$:

$$y - 3 = -\frac{3}{2}(x - (-4))$$ $$y - 3 = -\frac{3}{2}(x + 4)$$ $$y - 3 = -\frac{3}{2}x - \frac{3}{2} \times 4$$ $$y - 3 = -\frac{3}{2}x - 6$$ $$y = -\frac{3}{2}x - 6 + 3$$ $$y = -\frac{3}{2}x - 3$$

The equation of the diagonal PR is $\boxed{y = -\frac{3}{2}x - 3}$.

The equation of PR is $y = -\frac{3}{2}x - 3$.
The y-intercept occurs when $x=0$.

$$y = -\frac{3}{2}(0) - 3$$ $$y = -3$$

The y-intercept is $(0, -3)$.
According to the question, the y-intercept is vertex R.
The coordinates of point R are $\boxed{(0, -3)}$.

c) Point Q(-8, -4) is one of the vertices of the rhombus in (a) above. Obtain the equation of QS leaving it in the form $ax + by = c$, where a, b and c are integers. Hence state the coordinate of point S.

In a rhombus, the diagonals are perpendicular. The gradient of PR is $m_{PR} = -\frac{3}{2}$.
The gradient of QS, $m_{QS}$, is the negative reciprocal of $m_{PR}$.

$$m_{QS} = -\frac{1}{m_{PR}} = -\frac{1}{-\frac{3}{2}} = \frac{2}{3}$$

The line QS passes through $Q(-8, -4)$ and has a gradient of $\frac{2}{3}$. Using the point-slope form $y - y_1 = m(x - x_1)$:

$$y - (-4) = \frac{2}{3}(x - (-8))$$ $$y + 4 = \frac{2}{3}(x + 8)$$

Multiply by 3 to clear the fraction:

$$3(y + 4) = 2(x + 8)$$ $$3y + 12 = 2x + 16$$

Rearrange into the form $ax + by = c$:

$$2x - 3y = 12 - 16$$ $$2x - 3y = -4$$

The equation of QS is $\boxed{2x - 3y = -4}$.

The center of the rhombus is the intersection of the diagonals PR and QS.
Equation of PR: $y = -\frac{3}{2}x - 3$
Equation of QS: $2x - 3y = -4$
Substitute the expression for $y$ from the PR equation into the QS equation:

$$2x - 3\left(-\frac{3}{2}x - 3\right) = -4$$ $$2x + \frac{9}{2}x + 9 = -4$$

Multiply by 2 to eliminate the fraction:

$$4x + 9x + 18 = -8$$ $$13x = -8 - 18$$ $$13x = -26$$ $$x = -2$$

Substitute $x = -2$ back into the equation for PR to find $y$:

$$y = -\frac{3}{2}(-2) - 3$$ $$y = 3 - 3$$ $$y = 0$$

The center of the rhombus, let's call it $C$, is $(-2, 0)$.

The center $C(-2, 0)$ is the midpoint of the diagonal QS. Let $S = (x_S, y_S)$.
Using the midpoint formula $C = \left(\frac{x_Q+x_S}{2}, \frac{y_Q+y_S}{2}\right)$ with $Q(-8, -4)$:
For the x-coordinate:

$$-2 = \frac{-8 + x_S}{2}$$ $$-4 = -8 + x_S$$ $$x_S = -4 + 8$$ $$x_S = 4$$

For the y-coordinate:

$$0 = \frac{-4 + y_S}{2}$$ $$0 = -4 + y_S$$ $$y_S = 4$$

The coordinates of point S are $\boxed{(4, 4)}$.

From part (b), $R = (0, -3)$.
From part (c), $S = (4, 4)$.
The gradient $m_{RS}$ is given by $m = \frac{y_2-y_1}{x_2-x_1}$.

$$m_{RS} = \frac{4 - (-3)}{4 - 0}$$ $$m_{RS} = \frac{4 + 3}{4}$$ $$m_{RS} = \frac{7}{4}$$

The gradient of the line RS is $\boxed{\frac{7}{4}}$.