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.
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.