Are points (3, 2), (-2, -3), and (2, 3) form a triangle?

Let the given points be A(3,2), 8(-2,-3) and C(2,3)

Therefore

AB=$\sqrt{{(3+2)}^2+{(2+3)}^2} =\sqrt{50} =5\sqrt{2 }units$ 

BC=$\sqrt{{(-2-2)}^2+{(-3-3)}^2}=\sqrt{52}=2\sqrt{13 }units$

AC=$\sqrt{{(3-2)}^2+{(2-3)}^2} =\sqrt{2 }units$

Now, we can see that

${\left(2\sqrt{13}\right)}^2={\left(5\sqrt{2}\right)}^2+{\left(\sqrt{2}\right)}^2$

52 = 50+2

52 = 52

BC² = AB² = AC²

therefore, the given triangle is a right-angle triangle.