The points A(-1,2), B(5,2) and C(2,5) form an isosceles triangle because the angles CAB and CBA are congruent based on the tangent definition. The tangent of both angles is 3/3 since the height h from C to the line y=2 is 3 and the adjacent sides AD and DB are also both 3. The area of the isosceles triangle ABC is 9, which is calculated as 0.5 times the base AB of 6 times the height h of 3.

1. Show that the points A(-1,2), B(5,2) & C(2,5) are the vertices of an isosceles triangle. Find the
area of triangle ABC.
Solution
By definition, an isosceles triangle is one that has two congruent (equal) angles. To prove two
angles equal on the triangle ABC, let us consider the definition of tangent:
tanx = opposite/adjacent = h/a
Noting that the y values for points A and B are equal, the height is the distance from point C to
the line y = 2:
h = 5 - 2 = 3
The distance AB = 5 - -1 = 6
Let D be the point through C perpendicular to AB; i.e. the line x = 2.
AD = 2 - -1 = 3 ; DB = 5 - 2 = 3
(note that a triangle with perpendicular bisector is isosceles. But proof by definition follows)
The tangent of angle CAB = h/AD = 3/3 = 1
The tangent of angle CBA = h/DB = 3/3 = 1
Therefore the angles are congruent, and the triangle is isosceles.
The definition of the Area of a triangle is half base times height. We know both from above:
h = 3, AB = 6
0.5*h*AB = 9