How do you find the area of this triangle using determinant and vertices?
The instruction is : use a determinant and the vertices of the triangle to find the area of the triangle.
Vertices: (2,-2), (8,5), (6,-10)
The answer is 26. Does anyone know how to get the answers?
the answer to this problem is 26!!!! How can you get 26 as the answer?!!!!!
Use this formula(it uses the third-order determinant)
If the 3 points are
(x1,y1)
(x2,y2)
(x3,y3)
Then the area A is:
A =
|x1 y1 1|
|x2 y2 1|
|x3 y3 1|
^_^
you take the determinant of the matrix who first row is 1,2,-2;
whose second row is 1,8,5; and whose third row is 1,6,-10.
The first column has all 1’s. The second column is the set of x-values for the vertices, and the third column is the set of y-values. If you get a negative number, take the absolute value.
References :
Make determinants out of the vertices in pairs, in order:
|2 -2|…..|8.. 5|……..|6 -10|
|8. 5|…..|6 -10|……..|2..-2|….and find the values:
..26………-110………..8….then add and divide by 2.
(26-110+8)/2=-76/2=-38
If it comes out negative, drop the minus sign. That’s the area, 38.
References :
first get the distance
2,-2 /8,5
d= sqrt of 6^2+ 7^2
36+49= sqrt of 87
then 8,5 6-10
4+225= sqrt of 229
then 6,-10 2,-2
16+64= sqrt of 80
it’s a scalene triangle so use heron’s formula
sqrt of 80= 4 sqrt of 5
4 sqrt of 5+ sqrt of 229 + sqrt of 87 all over 2. that is the s.
sqrt of s(s-a)(s-b)(s-c)
when a,b,c are the sides.
References :
Use this formula(it uses the third-order determinant)
If the 3 points are
(x1,y1)
(x2,y2)
(x3,y3)
Then the area A is:
A =
|x1 y1 1|
|x2 y2 1|
|x3 y3 1|
^_^
References :