The line $L_1$ has equation $4y + 3 = 2x$ - Edexcel - A-Level Maths Pure - Question 9 - 2012 - Paper 2

Since $L_2$ is perpendicular to $L_1$, we find the slope of $L_1$.
Rearranging $4y + 3 = 2x$ gives:

$y = \frac{1}{2}x - \frac{3}{4}$
Thus, the slope $m_1 = \frac{1}{2}$. Therefore, the slope of $L_2$ is:

$m_2 = -\frac{1}{m_1} = -2$
Using point-slope form at point $C (2, 4)$, we get:

$y - 4 = -2(x - 2)$
Expanding this, we have:
$y - 4 = -2x + 4\\ y + 2x - 8 = 0$
This can be rearranged to give the equation for $L_2$:

$2x + y - 8 = 0$

To find intersection point $D$, set the equations of $L_1$ and $L_2$ equal.
Start with:

1. $4y + 3 = 2x$
2. $2x + y - 8 = 0$
   From equation 2, express $y$:
   $y = 8 - 2x$
   Substituting into equation 1:

$4(8 - 2x) + 3 = 2x$
Solving gives:
$32 - 8x + 3 = 2x$
$35 = 10x \\ x = 3.5$
Substitute back to find $y$:
$y = 8 - 2(3.5) = 1$
Thus, the coordinates of point $D$ are $(3.5, 1)$.

To find the length of segment $CD$, use the distance formula:

$CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(3.5 - 2)^2 + (1 - 4)^2}$
Calculating gives:
$CD = \sqrt{(1.5)^2 + (-3)^2} = \sqrt{2.25 + 9} = \sqrt{11.25} = \frac{3}{2} \sqrt{5}$

To find the area of quadrilateral $ACBE$, we need the coordinates of points $A$, $B$, $C$, and $E$.
Assuming area is calculated using triangles:
$Area = Area_{ABC} + Area_{ABE}$
Calculating each triangle area with base and height method.
Area of triangle $ABC$:

Using coordinates $(2, 4)$, $(3.5, 1)$, and $(p, 4)$ to find:
$Area_{ABC} = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) |$
For triangle $ABE$, apply similar methods.
Adding areas gives total area:
$Area_{ACBE} = 45$.