How to determine if a triangle can be drawn with the given points. Given $3$ points
$$(x_1, y_1), (x_2, y_2), (x_3, y_3),$$ how does one determine whether they are vertices of a triangle?
Thanks.
 A: 1: Find area of triangle formed by 3 points. if not zero they can form triangle.
2: Find line equation of 2 vertices and check if the 3rd vertices is present on this line. if not they can form triangle. Similarly Find the slope of the line joining points A and B,then find the slope of the line joining A and C.If they are same,then you can't draw a triangle.(@rah4927)
A: The area of a triangle is given by the formula $$|\frac12 det(\vec{AB}, \vec{AC})|$$
Let $$\vec{A}=(x_1,y_1), \vec{B}=(x_2,y_2), \vec{C}=(x_3, y_3)$$
Then $$\vec{AB}=(x_2-x_1, y_2-y_1), \vec{AC}=(x_3-x_1,y_3-y_1)$$
Then it must hold that $$|  
(x_2-x_1)(y_3-y_1)-(y_2-y_1)(x_3-x_1)| \neq 0$$
A: A triangle can be drawn with three points iff these three points are three different points.
EDIT: in order to be more precise in the core of my answer, and not in comment: three distincts colinear points form a degenerate triangle, but a triangle nonetheless. It is a geometric figure with three angles (they can be calculated, and not in the case where there are only one or two disctinct points) and their sum is 180°...
