In the diagram, A is the point $(-2, 4)$, and C is the point $(4, 7)$. AB is a vertical line. The line BC has gradient $\frac{3}{2}$.

1. Use the vertical line $AB$ to determine the $x$-coordinate of $B$.

Since $AB$ is vertical and passes through $A(-2,4)$, every point on $AB$ has $x=-2$.

So let $B=(-2,y)$.

2. Apply the gradient formula to line $BC$ to find the unknown $y$-coordinate.

Using $B=(-2,y)$ and $C=(4,7)$,

$\text{gradient of }BC=\frac{7-y}{4-(-2)}=\frac{3}{2}$

$\frac{7-y}{6}=\frac{3}{2}$

$7-y=9$

$y=-2$

3. State the coordinates of $B$.

Therefore,

$B=(-2,-2).$

1. Find the equation of line $BC$ using its gradient and point $C$.

Line $BC$ has gradient $\dfrac{3}{2}$ and passes through $C(4,7)$.

$y-7=\frac{3}{2}(x-4)$
$2y-14=3x-12$
$3x-2y+2=0$

2. Apply the perpendicular distance formula from point $A$ to line $BC$.

For a line $ax+by+c=0$, the perpendicular distance from $(x_1,y_1)$ to the line is

$\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}.$

Here, line $BC$ is $3x-2y+2=0$ and $A=(-2,4)$.

$\text{distance}=\frac{|3(-2)-2(4)+2|}{\sqrt{3^2+(-2)^2}}$

3. Calculate and state the shortest distance.

$\text{distance}=\frac{|-6-8+2|}{\sqrt{9+4}}$
$=\frac{12}{\sqrt{13}}$

However, since $B=(-2,-2)$ lies on $BC$ and $AB$ is vertical, the perpendicular distance is not along $AB$ unless $BC$ is horizontal. Checking by the formula gives the shortest distance:

$\frac{12}{\sqrt{13}} \approx 3.33$

1. Find the gradient of line $AC$ using the coordinates of $A$ and $C$.

For $A(-2,4)$ and $C(4,7)$,
$m_{AC}=\frac{7-4}{4-(-2)}=\frac{3}{6}=\frac{1}{2}.$

2. Rewrite the equation of line $L$ to identify its gradient and compare it with line $AC$.

For line $L$,
$\frac{4}{3}y-\frac{2}{3}x=9$
Multiply by $3$:
$4y-2x=27$
$4y=2x+27$
$y=\frac{1}{2}x+\frac{27}{4}.$
So the gradient of $L$ is $\frac{1}{2}$.

Since $m_L=m_{AC}=\frac{1}{2}$, the lines are parallel. Hence line $L$ does not intersect line $AC$.

1. Find the vertices of the reflected triangle and the relevant boundary lines.

Reflecting triangle $ABC$ about the $y$-axis changes each $x$-coordinate to its opposite.

$A(-2,4)\to P(2,4),\quad B(-2,-2)\to Q(2,-2),\quad C(4,7)\to R(-4,7).$

For triangle $ABC$:
$AC: y=\frac12x+5,\qquad BC: y=\frac32x+1.$

For the reflected triangle $PQR$:
$PR: y=-\frac12x+5,\qquad QR: y=-\frac32x+1.$

2. Identify the vertices of the overlapping region.

The overlapping region is bounded above by $AC$ and $PR$, and below by $BC$ and $QR$.

Top intersection:
$\frac12x+5=-\frac12x+5$
$x=0,\quad y=5.$
So the top vertex is $(0,5)$.

Bottom intersection:
$\frac32x+1=-\frac32x+1$
$x=0,\quad y=1.$
So the bottom vertex is $(0,1)$.

The left and right vertices are the touching points on the vertical sides:
$(-2,4)\quad \text{and}\quad (2,4).$

Hence the overlapping region has vertices
$(-2,4),\ (0,5),\ (2,4),\ (0,1).$

3. Calculate the area of the overlapping region.

The overlapping region is a kite with perpendicular diagonals.

Horizontal diagonal:
$2-(-2)=4.$

Vertical diagonal:
$5-1=4.$

Area of kite:
$\frac12 \times 4 \times 4=8.$