Find the area of the triangle with vertices , , and .
are you aware of Heron's formula?
Of course, there are other ways to approach this, like the (1/2)base*height formula
Let A(-1, -4) , B(3, -3), and C(-4, 4) be the coordinates of triangle. Please see attached diagram.
Using distance formula find
$\displaystyle BC = a = \sqrt{(3+4)^2+(-3-4)^2}= \sqrt{98}$
$\displaystyle CA = b = \sqrt{(-4+1)^2+(4+4)^2}= \sqrt{73}$
$\displaystyle AB = c = \sqrt{(-1-3)^2+(-4+3)^2}= \sqrt{17}$
Now, use cos law to find angle C
$\displaystyle \cos C = \frac{a^2+b^2-c^2}{2ab}$
$\displaystyle \cos C = \frac{98+73-17}{2\sqrt{98}\sqrt{73}}$
$\displaystyle \cos C = 0.910366$
C = 24.44 degrees
Area of triangle ABC $\displaystyle = \frac{1}{2}ab \;\sin C$
$\displaystyle = \frac{1}{2}\sqrt{98}\sqrt{73} \;\sin 24.44$
= 17.5 square units.