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.
 A: Enclose the quadrilateral in a rectangle $R$ with vertices of $(3,-5),(6,-5),(6,0),(3,0)$. This rectangle will have area $3\cdot 5 =15$ but will overshoot the quadrilateral in exactly four triangular areas.
Finding the area of these triangles should be straight forward, as their bases and heights are parallel the coordinate axes - I found two triangles of area $3$ and another two of area $1$, so all four together will have combined area $2(3)+2(1)=8$.
Take the difference:$15-8=7$.
A: 
you need to know determinants for using this. this is more easier and faster way to do it.
A: join diagonal AC. your quadrilateral will get divided into two triangles ABC 
ADB. find length of AC AB BC CD DA. and use herons formula to find area of two triangles. addition will give you the require area.
A: Hint 1
The area of quadrilateral $ABCD$ is equal to the sum of areas of triangles $ABC$ and $ACD$
Hint 2
$$P_{\triangle XYZ} = \frac{|\vec{XY}\times \vec{XZ}|}{2}  $$
Where $\times $ is the cross product of vectors:
$(a,b)\times (x,y) = (0,0,ay-bx)$
A: 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 $ABCD = \sqrt{\frac{145}{13}} \cdot \sqrt{13} = \sqrt{145}$
