Finding the area of a quadrilateral with two points given and other lie on a line (Coordinate Geometry) The question is as follows:
The diagram shows the quadrilateral $ABCD$ in which $A$ is the point $(4, 2)$ and $B$ is the point $(-2, -10)$. The points $C$ and $D$ lie on the line $x = 14$. The diagonal $AC$ is perpendicular to $AB$ and passes through the mid-point, $M$, of the diagonal $BD.$ Find the area of the quadrilateral $ABCD.$
I found $C$ and proceeded to try and find the mid-point, using colinear points. It came to nothing however. Can anyone on this forum help me?
 A: Since the midpoint of $BD$ lies on $AC$ we have $[ABCD]=2[ABC]$.$C$ is located at $(14;-3)$, so by the shoelace formula
$$ [ABCD]=\left|4\cdot(-10)+(-2)\cdot(-3)+14\cdot 2-2\cdot(-2)-(-10)\cdot 14-(-3)\cdot 4\right|=\color{red}{150} $$
and we don't even need to find where $D$ actually is.

Or we may locate $D$ at $(14;12)$ and deduce that
$[ABCD]=CD\cdot d(A,CD) = 15\cdot 10 = \color{red}{150}.$
A: Since the diagonal $AC$ is perpendicular to the diagonal $AB$ and $AB$ has slope $-\dfrac{4}{3},$ the diagonal $AC$ has slope $\dfrac{3}{4}.$ Since it contains the point $A (4,2),$ it has equation $y-2=\dfrac{3}{4}(x-4).$ So the $y$-coordinate of point $C$ is simply the value of $y$ when $x=14$ is substituted into this equation. This gives that $C$ is the point $(14,9.5).$ Let $D$ have coordinates $(14,a).$ Then the midpoint of $BD$ is $\left(6,5+\dfrac{a}{2}\right).$ Since the diagonal $AC$ passes through this point, we have that $3+\dfrac{a}{2}=\dfrac{3}{2}\Rightarrow a=-3.$ 
So the area is simply the area of the rectangle with width parallel to the $x$-axis equal to $16$ and with height parallel to the $y$-axis equal to $13$ minus the sum of the areas of the triangles and rectangle that are not included in the quadrilateral. This gives that the area is $16\cdot 13 - \dfrac{1}{2}\cdot16\cdot \dfrac{1}{2}- \dfrac{1}{2}\cdot8\cdot 4-\dfrac{1}{2}\cdot 5\cdot 10-5\cdot 4=\boxed{143.}$
