Question

A triangle has vertices at (-2,-3),(4, -3), and (3,5). What is the area of the triangle?

Answer 1

Answer:

this triangle is ABC with A(-2 ; -3), B(4; - 3) and C(3;5)

=> $AB=\sqrt{(4-(-2))^{2}+\sqrt{(-3 -(-3))^{2} } }=\sqrt{6^{2} }=6\\\\AC=\sqrt{(3-(-2))^{2}+(5-(-3))^{2} }=\sqrt{5^{2}+8^{2} }=\sqrt{89}\\\\BC =\sqrt{1^{2}+8^{2} }=\sqrt{65}$

using Heron theorem, we have:

$S=\sqrt{p(p-AB)(p-AC)(p-BC)}\\\\S=\sqrt{(\frac{6+\sqrt{89}+\sqrt{65} }{2})(\frac{\sqrt{89}+\sqrt{65}-6 }{2} )(\frac{6+\sqrt{65}-\sqrt{89} }{2})(\frac{6+\sqrt{89}-\sqrt{65} }{2}) }\\\\S=24$

with S is the area of the triangle

       $p=\frac{AB+AC+BC}{2}=\frac{6+\sqrt{89}+\sqrt{65} }{2}$

Step-by-step explanation: