Finding a vertex of a triangle knowing the other two and its area I have vertix A, vertix B and the area of a triangle, and I need to find the coordinates of vertex C, knowing that it's on the bisector between the first and the third sector of the Cartesian plane.
Until now I found out the length of the segment AB using the following formula:
$\sqrt{(x_b - x_a)^2 + (y_b - y_a)^2}$
Then I found out the mid-point of A e B and calculated the height of the triangle.
I now need to finally find the coordinates of the last vertex. How do I do this?
 A: The condition on vertex  $C$ is that it lies on the line $y=x$.
The area is known, and the length of the base $AB$ is known.  So the height $h$ with respect to this base is known.
Find the equation of the line $AB$. I am sure that you know how to do this. Suppose that this equation turns out to be $ax+by+c=0$. 
Since our point $C$ lies on the line $y=x$, we can suppose that its coordinates are $(t,t)$.  We will use the formula for the perpendicular distance from a point $(u,v)$ to the line $ax+by+c=0$. It is a standard fact that this distance is equal to 
$$\frac{|au+bv+c|}{\sqrt{a^2+b^2}}.$$
We know $a$, $b$, $c$, and the distance $h$.
So we need to solve the following equation for $t$:
$$\frac{|at+bt+c|}{\sqrt{a^2+b^2}}=h.$$
Or, removing absolute value signs, we need to solve the two linear equations
$$(a+b)t+c=h\sqrt{a^2+b^2}\quad\text{and}\quad (a+b)t+c=-h\sqrt{a^2+b^2}.$$
Remark: The idea is easily extended to solving similar problems, where $C$ lies on any line with given equation, like $2x+3y+4=0$ instead of $y=x$.
