Question

What is the area of triangle ABC with vertices A(2, 3), B(1, -3), and C(-3, 1)?

Answer 1

Area = 14 square units

Explanation:

First, after applying the distance formula `a^2+b^2=c^2`, we find that side length opposite to point A (call it `a`) `a=4sqrt2`, `b=sqrt29`, and `c=sqrt37`.
Next, use Herons rule:
`Area = sqrt(s(s-a)(s-b)(s-c))` where `s=(a+b+c)/2`.
We then get:
`Area = sqrt[(2sqrt2+1/2sqrt29+1/2sqrt37)(-2sqrt2+1/2sqrt29+1/2sqrt37)(2sqrt2-1/2sqrt29+1/2sqrt37)(2sqrt2+1/2sqrt29-1/2sqrt37)]`
It's not as scary as it looks. This simplifies down to:
`Area = sqrt196`, so `Area =14` `units^2`