Do the points $(3,2), (-2,-3)$ and $(2,3)$ form a triangle? If so,name the type of triangle formed.

(D) Let the points be $P(3,2), Q(-2,-3)$,and $R(2,3)$.
We use the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ to find the lengths of the sides:
$PQ = \sqrt{(-2-3)^2 + (-3-2)^2} = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25+25} = \sqrt{50} = 5\sqrt{2}$.
$QR = \sqrt{(2 - (-2))^2 + (3 - (-3))^2} = \sqrt{4^2 + 6^2} = \sqrt{16+36} = \sqrt{52} = 2\sqrt{13}$.
$PR = \sqrt{(2-3)^2 + (3-2)^2} = \sqrt{(-1)^2 + 1^2} = \sqrt{1+1} = \sqrt{2}$.
Since the sum of any two sides is greater than the third side ($PQ + PR > QR$,$PQ + QR > PR$,and $PR + QR > PQ$),the points form a triangle.
Checking for a right-angled triangle using the converse of the Pythagoras theorem: $PR^2 + PQ^2 = (\sqrt{2})^2 + (\sqrt{50})^2 = 2 + 50 = 52$.
Since $QR^2 = (\sqrt{52})^2 = 52$,we have $PR^2 + PQ^2 = QR^2$.
Thus,the triangle is a right-angled triangle.