SOLUTION: Given the points A(1,5), B(5,-2) C(-5,-5) and D(-15,1). 1.Find the area of the quadrilateral ABCD. Hint:divide the area into two triangles. 2.find the equation of a line AB?

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Question 1143813: Given the points A(1,5), B(5,-2) C(-5,-5) and D(-15,1).
1.Find the area of the quadrilateral ABCD. Hint:divide the area into two triangles.
2.find the equation of a line AB?
Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Given the points A(1,5), B(5,-2) C(-5,-5) and D(-15,1).
1.Find the area of the quadrilateral ABCD.
Here's a better way to do these:
     A    B    C    D    A
x    1    5   -5   -15   1
Y    5   -2   -4    1    5

---
Add the diagonal products starting at the upper left
1*-2 + 4*-4 + -5*1 + -15*5 = -2 -16 -5 -75 = -98
----
Add the diagonal products starting at the lower left
5*5 + -2*-5 + -4*-15 + 1*1 = 25 + 10 + 60 + 1 = 96
------
The difference between the 2 sums is 194
The area is 1/2 that: Area = 97 sq units
---------
This method works for all convex polygons with any # of vertices.
it might work for polygons that are not convex, IDK about that.
The points have to be in order around the polygon, tho.
//////////////////////
I tried a few non-convex polygons, and it does work.
\\\\\\\\\\\\\\\\\\\\\\
=====================================
Hint:divide the area into two triangles.
Hint: we don't need hints.
---------
2.find the equation of a line AB?

|x   y   1|
|1   5   1| = 0
|5  -2   1|

x*(5+2) - y*(1-5) + (-2-25) = 0
Answer by ikleyn(52900)   (Show Source): You can put this solution on YOUR website!
.

            The method to which Alan refers in his post is very robust,  but Alan made a number of errors on the way,
            so I came to fix it and to get the correct answer/solution.


Given the points A(1,5), B(5,-2) C(-5,-5) and D(-15,1).
1.Find the area of the quadrilateral ABCD.
Here's a better way to do these:
     A    B    C    D    A
x    1    5   -5   -15   1
Y    5   -2   -5    1    5       <----

---
Add the diagonal products starting at the upper left
1*(-2) + 5*(-5) + (-5)*1 + (-15)*5 = -2 -25 -5 -75 = -107     <----
----
Add the diagonal products starting at the lower left
5*5 + (-2)*(-5) + (-5)*(-15) + 1*1 = 25 + 10 + 75 + 1 = 111     <----
------
The difference between the 2 sums is 11 - (-107) = 111 + 107 = 218.             <----
The area is 1/2 that: Area = 218/2 = 109 sq units                 <----


Answer for the area :   109 square units.

By the arrows  " <---- "  I showed the lines where I edited Alan's calculations.

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

For this method,  see the site
https://www.mathopenref.com/coordpolygonarea.html

It works for non-convex polygons,  too.

It works,  practically,  for all and any polygons.

The major and the only restriction is  THIS :   a polygon should not have self-intersections.



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