<s>[INST] Find the equation of line $AP$ in triangle $ABC$ where $P$ will make $PA=PB=PC$ I was recently doing COMC (Canadian Opening Mathematic Challenge) past year exams. Here is another question that I want to ask!
Here is the question:
Triangle $ABC$ has its sides determinded in the following way: side $AB$ by line $3x-2y+3 = 0$, side $BC$ by line $x+y-14 = 0$, and side $AC$ by line $y=3$. If the point $P$ is chosen so that $PA=PB=PC$, determine the equation of the line containing $A$ and $P$
So I successfully figured out the three coordinates of the triangle $ABC$, which are (1,3), (5,9), (11,3). Now I can't find the point $P$ that makes $PA=PB=PC$. Hope someone would like to help me!
Thank you very much!
This question is from COMC 1998 Part B Question 1.
 [/INST] In other words, the point $P$
is the center of the circumscribed circle,
and
the coordinates of the circumcenter
is known to be found as
\begin{align}
P&=
\frac{a^2(b^2+c^2-a^2)\cdot A+b^2(a^2+c^2-b^2)\cdot B+c^2(b^2+a^2-c^2)\cdot C}
{a^2(b^2+c^2-a^2)+b^2(a^2+c^2-b^2)+c^2(b^2+a^2-c^2)}
\tag{1}\label{1}
,
\end{align}
where $a=|BC|$, $b=|AC|$, $c=|AB|$.
For $A=(1,3),\ B=(5,9),\ C=(11,3)$
the coordinates of the circumcenter
is therefore $P=(6,4)$,
and the question is simplified to:
find the equation of the line through
the two distinct points with known coordinates.
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