The circumcenter can be found as the intersection of the perpendicular bisectors.
The middle points of each one of the segments are
[tex]\begin{gathered} AB=(\frac{4+14}{2},\frac{12+6}{2})=(9,9) \\ BC=(\frac{14-6}{2},\frac{6+2}{2})=(4,4) \\ CA=(\frac{-6+4}{2},\frac{2+12}{2})=(-1,7) \\ \text{Where we used the formula} \\ M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) \end{gathered}[/tex]Furthermore, the slopes of each one of the three segments are
[tex]\begin{gathered} m_{AB}=\frac{6-12}{14-4}=-\frac{6}{10}=-\frac{3}{5} \\ m_{BC}=\frac{6-2}{14+6}=\frac{4}{20}=\frac{1}{5} \\ m_{CA}=\frac{2-12}{-6-4}=-\frac{10}{-10}=1 \\ \text{where we used the formula} \\ m=\frac{y_2-y_1}{x_2-x_1} \end{gathered}[/tex]In general, two lines are perpendicular if the product of their slopes is equal to -1. Then, the slopes of a perpendicular line to each one of the segments are
[tex]\begin{gathered} m^{\prime}_{AB}=\frac{5}{3} \\ m^{\prime}_{BC}=-5 \\ m^{\prime}_{CA}=-1 \end{gathered}[/tex]Given a point on a line and its slope, we can calculate the equation of any line. In our case,
[tex]\begin{gathered} bisectorofAB\colon y-9=\frac{5}{3}(x-9) \\ \Rightarrow bisectorofAB\colon y=\frac{5}{3}x-6 \\ bisectorofBC\colon y-4=-5(x-4)_{} \\ \Rightarrow bisectorofBC\colon y=-5x+24 \end{gathered}[/tex]Finally, calculate the intersection point of lines AB and BC as shown below
[tex]\begin{gathered} y=\frac{5}{3}x-6,y=-5x+24 \\ \Rightarrow\frac{5}{3}x-6=-5x+24 \\ \Rightarrow\frac{5}{3}x+5x=30 \\ \Rightarrow\frac{20}{3}x=30 \\ \Rightarrow x=\frac{9}{2} \\ \Rightarrow y=-5(\frac{9}{2})+24=\frac{3}{2} \\ \Rightarrow y=\frac{3}{2} \\ \Rightarrow(\frac{9}{2},\frac{3}{2})\to\text{circumcenter} \end{gathered}[/tex]The answer is (9/2,3/2)
[tex]\begin{cases}y=\frac{5x}{3}-6 \\ y=-5x+24\end{cases}[/tex]Remember that (properties of equality)
[tex]\begin{gathered} a=b\Rightarrow a+c=b+c \\ \text{and} \\ a=b\Rightarrow a\cdot c=b\cdot c \end{gathered}[/tex]From the system of equations,
[tex]\begin{gathered} y=y \\ \Rightarrow\frac{5x}{3}-6=-5x+24 \end{gathered}[/tex]Then, add +6 to both sides of the equation,
[tex]\begin{gathered} \frac{5x}{3}-6+6=-5x+24+6 \\ \Rightarrow\frac{5x}{3}=-5x+30 \end{gathered}[/tex]Add +5x to both sides of the equation,
[tex]\begin{gathered} \frac{5x}{3}+5x=-5x+30+5x \\ \Rightarrow\frac{20x}{3}=30 \end{gathered}[/tex]Multiply both sides of the equation by 3
[tex]\begin{gathered} \frac{20x}{3}=30 \\ \Rightarrow20x=90 \end{gathered}[/tex]Finally, multiply both sides of the equation by 1/20
[tex]\begin{gathered} 20x=90 \\ \Rightarrow\frac{1}{20}\cdot20x=\frac{1}{20}\cdot90 \\ \Rightarrow x=\frac{90}{20} \end{gathered}[/tex]There was no need to use negative reciprocals.
