Determining the position of a vertex of a triangle 
A triangle has a vertex $A$ at $(0,3)$, vertex $B$ at $(4,0)$, and vertex $C$ at $(x,5)$. If the area of the triangle is $8$, what is the value of $x$?

I did this problem out by using the formula of the given coordinates of the vertices. My answer was 2.66. 
 A: The area of the triangle with vertices $(x_{i},y_{i})$, $i=1,2,3$ is given by $$\Delta = \frac{1}{2} \cdot \left| \begin{array}{cc} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\\ x_{3} & y_{3} & 1 \end{array}\right| = \frac{1}{2}\left|\begin{array}{cc} 0 & 3 & 1 \\  4 & 0 & 1 \\ x & 5 & 1 \end{array}\right|$$
Now substitute $\Delta=8$, and the corresponding values for $(x_{i},y_{i})$ and solve for $x$. Solving you get $16 = -3(4-x) + 1 \times 20 = -12 + 3x +20$. Therefore you have $3x=8$ which says $x=\frac{\pm{8}}{3}$. You get the minus sign if you interchange the vertices. (Thanks to Arturo for pointing out the error.)
A: Good work, in arriving at one of the correct solutions, Briana..  
I would just like to clarify what others mean by there being two solutions: there are two triangles, each sharing vertices at $(0,3)$ and $(4,0)$ (and hence they share a side).  But there are two values for the unknown $x$ in the vertex $(x, 5)$ that yield a triangle with area $8: x = \frac {8}{3} = (2.66$ repeating), and $x = -8.$  
In other words, the area of the triangle with third vertex $( \frac {8}{3}, 5)$ and the area of triangle with third vertex $(-8, 5)$ both equal $8$.

