To find area of quadrilateral $ABCD$ using the given co-ordinates. I am unable to get the answer to this question. The question is to find the area of a quadrilateral having its vertices as coordinates in order: 
$A(3,-2)$; $B(4,0)$; $C(6,-3)$ and $D(5,-5)$. 
I divided this into 2 triangles - $\Delta ABC$ and $\Delta BCD$. Now by the area of triangle formula, I am getting the total area to be $35$ square units but my tuition partner is getting a very different answer $7$ square units to be precise.

Here's a graph so you can see what is going on visually.

By looking at the graph, or explicitly finding the equations of the lines $BC$, $AD$, $CD$, $AB$ we can see that $AB$ is parallel to $AD$ and $CD$. is parallel  to $AB$.
Note that also $|BC| = |CD|$ and $|AB|=|CD|$.
Then we can deduce this quadrilateral is a parallelogram. The area is then equal to the product of the base and its height. That is the product of $AD$ and the perpendicular distance between $AD$ and $BC$.The equation of the line $AD$ is given by $y =-\frac{3}{2}x + \frac{5}{2}$
The gradient equation of the normal to this line is then given by $\frac{-1}{-\frac{3}{2}} = \frac{2}{3}$
This normal starts has a point on $AD$ and a corresponding point on $BC$
The equation of the normal is then given by $y = \frac{2}{3} x -4$
The equation of the line $BC$ is given by $y= -\frac{3}{2}x + 6$
Set these two equations equal to find the point of intersection:
$-\frac{3}{2}x + 6 = \frac{2}{3}x -4 \Rightarrow x =\frac{60}{13}, y = -\frac{12}{13}$
Let's call this point of intersection on $BC$ point $ E = (\frac{60}{13},-\frac{12}{13})$. 
Then we have $|AE| = \sqrt{(\frac{60}{13} - 3)^{2} + (\frac{12}{13} - - 2)^{2}} = \sqrt{\frac{145}{13}}$
$|AD| = \sqrt{(3-5)^{2}+(-2+5)^{2}} = \sqrt{13}$
Then the area of $ABCD = \sqrt{\frac{145}{13}} \cdot \sqrt{13} = \sqrt{145}$
