Re: vertices of a triangle

(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.