area of a triangle with points at (-9,5), (6,10), and (2,-10)

First we are going to draw the triangle using the given coordinates. 
Next, we are going to use the distance formula to find the sides of our triangle.
Distance formula: [tex]d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2} [/tex]

Distance from point A to point B:
[tex]d_{AB}= \sqrt{[6-(-9)]^2+(10-5)^2} [/tex]
[tex]d_{AB}= \sqrt{(6+9)^2+(10-5)^2} [/tex]
[tex]d_{AB}= \sqrt{(15)^2+(5)^2}[/tex]
[tex]d_{AB}= \sqrt{225+25}[/tex]
[tex]d_{AB}= \sqrt{250}[/tex]
[tex]d_{AB}=15.81[/tex]

Distance from point A to point C:
[tex]d_{AC}= \sqrt{[2-(-9)]^2+(-10-5)^2}[/tex]
[tex]d_{AC}= \sqrt{(2+9)^2+(-10-5)^2}[/tex]
[tex]d_{AC}= \sqrt{11^2+(-15)^2}[/tex]
[tex]d_{AC}= \sqrt{121+225}[/tex]
[tex]d_{AC}= \sqrt{346}[/tex]
[tex]d_{AC}= 18.60[/tex]

Distance from point B from point C
[tex]d_{BC}= \sqrt{(2-6)^2+(-10-10)^2}[/tex]
[tex]d_{BC}= \sqrt{(-4)^2+(-20)^2}[/tex]
[tex]d_{BC}= \sqrt{16+400}[/tex]
[tex]d_{BC}= \sqrt{416}[/tex]
[tex]d_{BC}=20.40[/tex]

Now, we are going to find the semi-perimeter of our triangle using the semi-perimeter formula:
[tex]s= \frac{AB+AC+BC}{2} [/tex]
[tex]s= \frac{15.81+18.60+20.40}{2} [/tex]
[tex]s= \frac{54.81}{2} [/tex]
[tex]s=27.41[/tex]

Finally, to find the area of our triangle, we are going to use Heron's formula:
[tex]A= \sqrt{s(s-AB)(s-AC)(s-BC)} [/tex]
[tex]A=\sqrt{27.41(27.41-15.81)(27.41-18.60)(27.41-20.40)} [/tex]
[tex]A= \sqrt{27.41(11.6)(8.81)(7.01)} [/tex]
[tex]A=140.13[/tex]

We can conclude that the perimeter of our triangle is 140.13 square units.