Official Solution: Given two points \(A(-2, 2)\) and \(B(3, 2)\). Question asks to find the area of triangle \(ABC\), where \(C(x,y)\). Look at the diagram below:
(1) \(|y-2|=1\). Either \(y=3\) or \(y=1\), hence vertex \(C\) could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the area 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 \(C\): \(C_1\) and \(C_2\), the heights of \(ABC_1\) and \(ABC_2\) are the same and equal to 1). So, we have that \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.
(2) Angle at the vertex \((x, y)\) equals 90 degrees. This statement says that \(ABC\) is a right triangle with hypotenuse \(AB\): consider \(AB\) to be diameter of a circle. 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 \(C\): \(C_3\) and \(C_4\), heights of \(ABC_3\) and \(ABC_4\) are different). Not sufficient.
Answer: A
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