The diagram shows a trapezium PQRS in which PQ is parallel to SR. The point P is (-1, -2) and the point Q is (1, 4).

1. Find the midpoint of PQ

Midpoint of $PQ = \left(\frac{-1 + 1}{2}, \frac{-2 + 4}{2}\right) = (0, 1)$

2. Find the gradient of PQ

Gradient of $PQ = \frac{4 - (-2)}{1 - (-1)} = \frac{6}{2} = 3$

3. Find the gradient of the perpendicular bisector

Gradient of perpendicular bisector $= -\frac{1}{3}$

4. Find the equation of the perpendicular bisector

Using $y - y_1 = m(x - x_1)$ with point $(0, 1)$ and gradient $-\frac{1}{3}$:
$y - 1 = -\frac{1}{3}(x - 0)$
$y = -\frac{1}{3}x + 1$

1. Find the equation of the perpendicular bisector of $PQ$ (from part i).

Midpoint of $PQ = \left(\frac{-1+1}{2}, \frac{-2+4}{2}\right) = (0, 1)$
Gradient of $PQ = \frac{4 - (-2)}{1 - (-1)} = \frac{6}{2} = 3$
Gradient of perpendicular bisector $= -\frac{1}{3}$
Equation of perpendicular bisector: $y - 1 = -\frac{1}{3}(x - 0) \implies y = -\frac{1}{3}x + 1$

2. Find the coordinates of $R$ by setting $y = 0$.

Since $R$ lies on the $x$-axis, $y = 0$.
$0 = -\frac{1}{3}x + 1$
$\frac{1}{3}x = 1 \implies x = 3$
The coordinates of $R$ are $(3, 0)$.

1. Determine the $y$-coordinate of $S$

Since $PS$ is parallel to the $x$-axis, the $y$-coordinate of $S$ is the same as the $y$-coordinate of $P$.
$y = -2$

2. Find the $x$-coordinate of $S$ using the equation of $RS$

Substitute $y = -2$ into the equation of $RS$:
$-2 = 3x - 9$
$3x = 7$
$x = \frac{7}{3}$

1. State the coordinates of all four vertices of the trapezium from earlier parts

From earlier parts:
$P = (-1, -2)$
$Q = (1, 4)$
$R = (3, 0)$
$S = \left(\frac{7}{3}, -2\right)$

2. Apply the shoelace formula to find the area of the trapezium

Arranging the vertices in counter-clockwise order $P, S, R, Q$:
$\text{Area} = \frac{1}{2} \left| \begin{matrix} -1 & \frac{7}{3} & 3 & 1 & -1 \\ -2 & -2 & 0 & 4 & -2 \end{matrix} \right|$
$\text{Area} = \frac{1}{2} \left| [(-1)(-2) + \left(\frac{7}{3}\right)(0) + (3)(4) + (1)(-2)] - [(-2)\left(\frac{7}{3}\right) + (-2)(3) + (0)(1) + (4)(-1)] \right|$
$\text{Area} = \frac{1}{2} \left| [2 + 0 + 12 - 2] - \left[-\frac{14}{3} - 6 + 0 - 4\right] \right|$
$\text{Area} = \frac{1}{2} \left| 12 - \left(-\frac{44}{3}\right) \right|$
$\text{Area} = \frac{1}{2} \left( \frac{36}{3} + \frac{44}{3} \right) = \frac{1}{2} \left( \frac{80}{3} \right) = \frac{40}{3}$