87.6k views
2 votes
Decide!!!!!!!!!!!!!!

Decide!!!!!!!!!!!!!!-example-1
User Snowindy
by
8.0k points

1 Answer

3 votes

A third alternative to the formulas I suggested in a comment is to use the vector cross product. Recast the three given points as points in 3D space with 0 z-coordinate:

A(20, 21, 0), B(8, 29, 0), and C(13, 1, 0)

Now consider two vectors,

u = B - A = 〈8, 29, 0〉 - 〈20, 21, 0〉 = 〈-12, 8, 0〉

v = C - A = 〈13, 1, 0〉 - 〈20, 21, 0〉 = 〈-7, -20, 0〉

Then recall the cross product identity,

||u × v|| = ||u|| ||v|| |sin(θ)|

where θ is the angle between the two vectors. The magnitude of u × v is equal to the area of the parallelogram spanned by u and v, and hence twice the area of the triangle of interest.

So we have

area = ||1/2 〈-12, 8, 0〉 × 〈-7, -20, 0〉||

area = 1/2 ||〈-12, 8, 0〉 × 〈-7, -20, 0〉||

area = 1/2 ||〈0, 0, 296〉||

area = 296/2

area = 148

User Rasmus Hansen
by
7.6k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories