Vertices of a traingle are at (-2,1) , (3,1) & (x,y) on a

Below is almost identical question:

Vertices of a triangle have coordinates (-2, 2), (3, 2), (x, y). What is the area of the triangle?

(1) |y - 2| = 1
(2) angle at the vertex \((x, y)\) equals 90 degrees

Given two points A(-2,2) and B(3,2). Question: Area ABC=?, where C(x,y).
(1) |y-2|=1 --> \(y=3\) or \(y=1\) --> vertex C could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the are of ABC will be the same --> \(area=\frac{1}{2}*base*height\) so \(base=AB=5\) and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.

(2) angle at the vertex C(x,y) equals to 90 degrees --> ABC is a right triangle with hypotenuse AB --> consider AB to be diameter of a circle and in this case C could be anywhere on the circle and it will be right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle), thus height of ABC will be different for different location of point C, resulting the different areas (see two possible locations of of C: C3 and C4, heights of ABC3 and ABC4 are different). Not sufficient.

Answer: A.

Hope it heps.