|
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(52887)  (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.
|
|
|
| |
