图示中，$OPQR$ 是一个梯形，其中 $OP$ 平行于 $RQ$，且 $OP$ 垂直于 $PQ$。 点 $P$ 的坐标为 $(-2, 6)$，且 $R$ 落在 $x$ 轴上。

1. Find the midpoint of OP

Midpoint of $OP = \left(\frac{0 + (-2)}{2}, \frac{0 + 6}{2}\right) = (-1, 3)$

2. Find the gradient of OP and the gradient of the perpendicular bisector

Gradient of $OP = \frac{6 - 0}{-2 - 0} = -3$
Gradient of perpendicular bisector $= -\frac{1}{-3} = \frac{1}{3}$

3. Find the equation of the perpendicular bisector

Using $y - y_1 = m(x - x_1)$ with point $(-1, 3)$ and gradient $\frac{1}{3}$:
$y - 3 = \frac{1}{3}(x - (-1))$
$y - 3 = \frac{1}{3}x + \frac{1}{3}$
$y = \frac{1}{3}x + \frac{10}{3}$

1. Find the coordinates of $R$ using the equation of the perpendicular bisector of $OP$.

From part (a), the equation of the perpendicular bisector of $OP$ is $y = \frac{1}{3}x + \frac{10}{3}$.
Since $R$ lies on the $x$-axis, $y = 0$.
$0 = \frac{1}{3}x + \frac{10}{3}$
$\frac{1}{3}x = -\frac{10}{3} \implies x = -10$
So, $R$ is $(-10, 0)$.

2. Find the equation of the line $RQ$.

Since $OP$ is parallel to $RQ$, the gradient of $RQ$ is the same as the gradient of $OP$.
Gradient of $OP = \frac{6 - 0}{-2 - 0} = -3$.
Equation of $RQ$ passing through $R(-10, 0)$:
$y - 0 = -3(x - (-10))$
$y = -3x - 30$

3. Find the equation of the line $PQ$.

Since $OP$ is perpendicular to $PQ$, the gradient of $PQ$ is $m_{PQ} = -\frac{1}{-3} = \frac{1}{3}$.
Equation of $PQ$ passing through $P(-2, 6)$:
$y - 6 = \frac{1}{3}(x - (-2))$
$y = \frac{1}{3}x + \frac{2}{3} + 6$
$y = \frac{1}{3}x + \frac{20}{3}$

4. Solve the simultaneous equations for $RQ$ and $PQ$ to find the coordinates of $Q$.

Equating the two expressions for $y$:
$-3x - 30 = \frac{1}{3}x + \frac{20}{3}$
Multiply by 3:
$-9x - 90 = x + 20$
$-10x = 110 \implies x = -11$
Substitute $x = -11$ into $y = -3x - 30$:
$y = -3(-11) - 30 = 33 - 30 = 3$
Therefore, the coordinates of $Q$ are $(-11, 3)$ (shown).

1. Find the coordinates of $R$ using information from earlier parts

From part (a), the perpendicular bisector of $OP$ has equation $y = \frac{1}{3}x + \frac{10}{3}$.
Since $R$ lies on the $x$-axis and on this bisector, substitute $y = 0$:
$0 = \frac{1}{3}x + \frac{10}{3} \implies x = -10$
So, $R$ is $(-10, 0)$.

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

The vertices of the trapezium in anticlockwise order are $O(0, 0)$, $R(-10, 0)$, $Q(-11, 3)$, and $P(-2, 6)$.
$\text{Area} = \frac{1}{2} \left| \begin{matrix} 0 & -10 & -11 & -2 & 0 \\ 0 & 0 & 3 & 6 & 0 \end{matrix} \right|$
$\text{Area} = \frac{1}{2} | (0)(0) + (-10)(3) + (-11)(6) + (-2)(0) - [(0)(-10) + (0)(-11) + (3)(-2) + (6)(0)] |$
$\text{Area} = \frac{1}{2} | -30 - 66 - (-6) |$
$\text{Area} = \frac{1}{2} | -90 | = 45$