The triangle with vertices at (1,0), (0,1), and (k,k) has area 1/2. Find k.
Obviously, \(k=0 \) is an answer. Let's try to find some more (or prove that k=0 is the only answer). Let the base of the triangle be the segment from (1, 0) to (0, 1). The length of this segment will then be \(\sqrt{2}\). We can see that the height of the triangle be \(\frac{\sqrt{2}}{2}\). Since the third point is (k, k), the point must lie on the line x=y which means that k can be 0 or 1.