# 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?

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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 $y=x$.
 I'm looking for the fastest way, so if you could explain that one too I would appreciate it :) – user1301428 May 4 '12 at 17:40 Thanks for this reply too. So, I have (for now :D) a couple of questions about your original solution: 1) isn't the equation of the bisector $y = x$? 2) How can you say that the bisector of the Cartesian plane is perpendicular to $AB$? – user1301428 May 4 '12 at 18:06 Maybe my question is a bit confusing, but with bisector I mean the bisector of the first and third sector of the Cartesian plane, not of a specific segment. Imagine having $A = (1,2)$ and $B = (4,0)$, in this case this bisector is not perpendicular to $AB$. – user1301428 May 4 '12 at 20:09 Exactly! Sorry if the question was confusing. – user1301428 May 4 '12 at 20:16 Thank you for improving the answer :) – user1301428 May 5 '12 at 12:08