The diagram shows lines $l_1$ and $l_2$ intersecting at $P$. Line $l_1$ has equation $3y - 4x + 40 = 0$ and $l_2$ has equation $4y + x = 48$. The point $M$ is the midpoint of $OP$. The line $l_2$ intersects the $y$-axis at $Q$ and the line $l_1$ intersects the $x$-axis at $R$.

1. Find the coordinates of $P$ by solving the simultaneous equations for $l_1$ and $l_2$.

From $l_2$, $x = 48 - 4y$.
Substitute into $l_1$:
$3y - 4(48 - 4y) + 40 = 0$
$3y - 192 + 16y + 40 = 0$
$19y = 152$
$y = 8$
Substitute $y = 8$ back to find $x$:
$x = 48 - 4(8) = 16$
So, $P$ is $(16, 8)$.

2. Find the coordinates of $M$, the midpoint of $OP$.

$M = \left( \frac{16 + 0}{2}, \frac{8 + 0}{2} \right) = (8, 4)$

3. Find the coordinates of $R$, the $x$-intercept of $l_1$.

Substitute $y = 0$ into $l_1$:
$3(0) - 4x + 40 = 0$
$4x = 40$
$x = 10$
So, $R$ is $(10, 0)$.

4. Calculate the gradients of $OP$ and $MR$ and show their product is $-1$.

Gradient of $OP$, $m_{OP} = \frac{8 - 0}{16 - 0} = \frac{1}{2}$
Gradient of $MR$, $m_{MR} = \frac{0 - 4}{10 - 8} = \frac{-4}{2} = -2$
Product of gradients $= m_{OP} \times m_{MR} = \frac{1}{2} \times (-2) = -1$
Since the product of their gradients is $-1$, $OP$ is perpendicular to $MR$.

1. Find the coordinates of the intersection point $P$ by solving the simultaneous equations for $l_1$ and $l_2$.

From $l_2$: $x = 48 - 4y$
Substitute into $l_1$:
$3y - 4(48 - 4y) + 40 = 0$
$3y - 192 + 16y + 40 = 0$
$19y = 152 \implies y = 8$
$x = 48 - 4(8) = 16$
So, $P = (16, 8)$.

2. Find the coordinates of $Q$ and $R$.

For $Q$ (on $y$-axis), substitute $x = 0$ into $l_2$:
$4y + 0 = 48 \implies y = 12$
So, $Q = (0, 12)$.

For $R$ (on $x$-axis), substitute $y = 0$ into $l_1$:
$3(0) - 4x + 40 = 0 \implies 4x = 40 \implies x = 10$
So, $R = (10, 0)$.

3. Calculate the area of quadrilateral $ORPQ$ using the shoelace formula.

The vertices in counter-clockwise order are $O(0,0)$, $R(10,0)$, $P(16,8)$, and $Q(0,12)$.
$\text{Area} = \frac{1}{2} \left| \begin{matrix} 0 & 10 & 16 & 0 & 0 \\ 0 & 0 & 8 & 12 & 0 \end{matrix} \right|$
$\text{Area} = \frac{1}{2} |(0 \times 0 + 10 \times 8 + 16 \times 12 + 0 \times 0) - (0 \times 10 + 0 \times 16 + 8 \times 0 + 12 \times 0)|$
$\text{Area} = \frac{1}{2} |80 + 192 - 0|$
$\text{Area} = \frac{1}{2} (272) = 136$