The diagram shows the quadrilateral $ABCD$ in which point $A$ is $(-1, 2)$ and point $C$ is $(8.5, 4)$. The point $P(6.5, 8)$ lies on $AD$ such that $AP : PD = 3 : 2$. The midpoint of $BC$, point $M$ lies on the $x$-axis and directly below point $P$.

1. Find the coordinates of $M$.

Since $M$ lies directly below $P(6.5, 8)$, its $x$-coordinate is $6.5$.
Since $M$ lies on the $x$-axis, its $y$-coordinate is $0$.
Therefore, $M = (6.5, 0)$.

2. Find the coordinates of $D$ using the given ratio $AP : PD = 3 : 2$.

The horizontal displacement from $A(-1, 2)$ to $P(6.5, 8)$ is $6.5 - (-1) = 7.5$.
This corresponds to 3 parts of the ratio.
1 part of horizontal displacement $= \frac{7.5}{3} = 2.5$.
The horizontal displacement from $P$ to $D$ (2 parts) is $2 \times 2.5 = 5$.
$x$-coordinate of $D = 6.5 + 5 = 11.5$.

The vertical displacement from $A$ to $P$ is $8 - 2 = 6$.
1 part of vertical displacement $= \frac{6}{3} = 2$.
The vertical displacement from $P$ to $D$ (2 parts) is $2 \times 2 = 4$.
$y$-coordinate of $D = 8 + 4 = 12$.

Therefore, $D = (11.5, 12)$.

3. Find the coordinates of $B$ using the midpoint formula.

Let the coordinates of $B$ be $(x, y)$.
Since $M(6.5, 0)$ is the midpoint of $BC$ and $C$ is $(8.5, 4)$:
$\frac{x + 8.5}{2} = 6.5 \implies x + 8.5 = 13 \implies x = 4.5$
$\frac{y + 4}{2} = 0 \implies y + 4 = 0 \implies y = -4$

Therefore, $B = (4.5, -4)$.

1. Find the coordinates of D and B from part (a)

Let $D = (x_D, y_D)$. Since $AP : PD = 3 : 2$, $P$ divides $AD$ in ratio $3:2$.
$\frac{2(-1) + 3x_D}{5} = 6.5 \implies -2 + 3x_D = 32.5 \implies 3x_D = 34.5 \implies x_D = 11.5$
$\frac{2(2) + 3y_D}{5} = 8 \implies 4 + 3y_D = 40 \implies 3y_D = 36 \implies y_D = 12$
So $D(11.5, 12)$.
$M$ is on the $x$-axis directly below $P(6.5, 8)$, so $M(6.5, 0)$.
$M$ is the midpoint of $BC$, so $\frac{x_B + 8.5}{2} = 6.5 \implies x_B = 13 - 8.5 = 4.5$
$\frac{y_B + 4}{2} = 0 \implies y_B = -4$
So $B(4.5, -4)$.

2. Apply the shoelace formula to find the area of quadrilateral ABCD

Vertices in anti-clockwise order: $A(-1, 2)$, $B(4.5, -4)$, $C(8.5, 4)$, $D(11.5, 12)$.
Area $= \frac{1}{2} \left| \begin{matrix} -1 & 4.5 & 8.5 & 11.5 & -1 \\ 2 & -4 & 4 & 12 & 2 \end{matrix} \right|$
$= \frac{1}{2} | ((-1)(-4) + (4.5)(4) + (8.5)(12) + (11.5)(2)) - ((2)(4.5) + (-4)(8.5) + (4)(11.5) + (12)(-1)) |$
$= \frac{1}{2} | (4 + 18 + 102 + 23) - (9 - 34 + 46 - 12) |$
$= \frac{1}{2} | 147 - 9 |$
$= \frac{1}{2} (138) = 69$