Since the polygon is convex (has no "sunk-in" places), we make this
determinant of coordinates of vertices, repeating the first one at
the bottom. [It doesn't matter which vertex we start with].



We multiply and add all the "northwest to southeast"-diagonals:

(-4)(-4)+(3)(2)+(6)(4)+(1)(2) = 16+6+24+2=48


We multiply and add all the "northeast to southwest"-diagonals:

(2)(3)+(-4)(6)+(2)(1)+(4)(-4) = 6-24+2-16=-32

Subtract: 48-(-32) = 48+32 = 80

Do the other one the same way.  Graph it to make sure it is
a convex polygon.  [If the polygon is concave, we would break it 
into convex polygons]. If the area comes out negative, just take
the absolute value, making it positive.

Edwin

Area of the polygon =  the POSITIVE difference of the SUM of the POSITIVE and NEGATIVE DIAGONAL-PRODUCTS.

First of all, you have to make sure that the points have been aligned in a CLOCKWISE or COUNTERCLOCKWISE position. 

Sum of POSITIVE DIAGONAL-PRODUCTS: - 32

Sum of NEGATIVE DIAGONAL-PRODUCTS: 48

 

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
 .

Since you have two different answers from two tutors, I came to bring my THIRD solution 

    (it is a joke, of course : I came to check them).


I consider part (a) only.


In this quadrilateral, we have horizontal diagonal from (-4,2) to (6,2).

It divides the quadrilateral in two triangles.

Their base has the length  10 = 6 - (-4)  units.


The upper triangle has the altitude (the height) of 2 = 4 - 2 units.
So, its area is   = 10 square units.


The lower triangle has the altitude 6 = 2 - (-4) units.
So, its area is   = 30 square units.


Hence, the entire area of the quadrilateral is  10 + 30 = 40 units.


It is the ANSWER  to question (a).






--------------





May I give a fair advise to you ?





    In the future, NEVER place more than one problem/question to your post.

    It ALWAYS works AGAINST your interests.






 

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