How to  graph this? I have a non-right triangle. I will call 


the bottom or base edge $b$,
    the top left edge $a$,
    the top right edge $c$.


Let $ a=c+2 $ and $b=10$. How do I graph a curve where the graph $ x  $ and $ y$ coordinates represent the vertex where edges $ a $ and $c$  meet?

Taking the origin as the intersection of $a$ and $b$, with the equation in terms of the angle $\theta$ at that vertex:
Through the cosine law we have:
$$(a-2)^2=a^2+100-20a\cos\theta$$
$$-4a+4=100-20a\cos\theta$$
$$a=\frac {96}{20\cos\theta-4}$$
With our particular setup, $a$ is the radius. So this is a polar equation, where
$$r(\theta)=\frac{96}{20\cos\theta-4}=\frac{24}{5\cos\theta-1}$$
we can convert to cartesian coordinates if you want:
$$5r\cos\theta-r=24$$
$$(5r\cos\theta-24)^2=r^2$$
$$(5x-24)^2=x^2+y^2$$
$$y^2=24x^2-240x+576$$
$$y^2=24(x-5)^2-24$$
$$(x-5)^2-\frac{y^2}{24}=1$$
which is a hyperbola. You want the rightmost branch (so $a>0$), which is valid when $\cos\theta<\frac 1 5$.
EDIT: picture time!

