The given triangle has vertices at:
[tex]\begin{gathered} P(-2,5) \\ Q(4,1) \\ R(-2,-3) \end{gathered}[/tex]In the coordinate plane, the triangle looks like this:
There are different forms to find the circumcenter, we are going to use the midpoint formula:
[tex]M(x,y)=(\frac{x1+x2}{2},\frac{y1+y2}{2})[/tex]Apply this formula for each vertice and find the midpoint:
[tex]M_{P,Q}=(\frac{-2+4}{2},\frac{5+1}{2})=(1,3)[/tex]For QR:
[tex]M_{Q,R}=(\frac{4+(-2)}{2},\frac{1+(-3)}{2})=(1,-1)[/tex]For PR:
[tex]M_{P,R}=(\frac{-2+(-2)}{2},\frac{5+(-3)}{2})=(-2,1)[/tex]Now, we need to find the slope for any of the line segments, for example, PQ:
We can apply the slope formula:
[tex]m=\frac{y2-y1}{x2-x1}=\frac{1-5}{4-(-2)}=\frac{-4}{6}=-\frac{2}{3}[/tex]By using the midpoint and the slope of the perpendicular line, find out the equation of the perpendicular bisector line, The slope of the perpendicular line is given by the formula:
[tex]\begin{gathered} m1\cdot m2=-1 \\ m2=-\frac{1}{m1} \\ m2=-\frac{1}{-\frac{2}{3}}=\frac{3}{2}_{} \end{gathered}[/tex]The slope-intercept form of the equation is y=mx+b. Replace the slope of the perpendicular bisector and the coordinates of the midpoint to find b:
[tex]\begin{gathered} 3=\frac{3}{2}\cdot1+b \\ 3-\frac{3}{2}=b \\ b=\frac{3\cdot2-1\cdot3}{2}=\frac{6-3}{2} \\ b=\frac{3}{2} \end{gathered}[/tex]Thus, the equation of the perpendicular bisector of PQ is:
[tex]y=\frac{3}{2}x+\frac{3}{2}[/tex]If we graph this bisector over the triangle we obtain:
Now, let's find the slope of the line segment QR:
[tex]m=\frac{-3-1}{-2-4}=\frac{-4}{-6}=\frac{2}{3}[/tex]The slope of the perpendicular bisector is:
[tex]m2=-\frac{1}{m1}=-\frac{1}{\frac{2}{3}}=-\frac{3}{2}[/tex]Let's find the slope-intercept equation of this bisector:
[tex]\begin{gathered} -1=-\frac{3}{2}\cdot1+b \\ -1+\frac{3}{2}=b \\ b=\frac{-1\cdot2+1\cdot3}{2}=\frac{-2+3}{2} \\ b=\frac{1}{2} \end{gathered}[/tex]Thus, the equation is:
[tex]y=-\frac{3}{2}x+\frac{1}{2}[/tex]This bisector in the graph looks like this:
Now, to find the circumcenter we have to equal both equations, and solve for x:
[tex]\begin{gathered} \frac{3}{2}x+\frac{3}{2}=-\frac{3}{2}x+\frac{1}{2} \\ \text{Add 3/2x to both sides} \\ \frac{3}{2}x+\frac{3}{2}+\frac{3}{2}x=-\frac{3}{2}x+\frac{1}{2}+\frac{3}{2}x \\ \frac{6}{2}x+\frac{3}{2}=\frac{1}{2} \\ \text{Subtract 3/2 from both sides} \\ \frac{6}{2}x+\frac{3}{2}-\frac{3}{2}=\frac{1}{2}-\frac{3}{2} \\ \frac{6}{2}x=-\frac{2}{2} \\ 3x=-1 \\ x=-\frac{1}{3} \end{gathered}[/tex]Now replace x in one of the equations and solve for y:
[tex]\begin{gathered} y=-\frac{3}{2}\cdot(-\frac{1}{3})+\frac{1}{2} \\ y=\frac{1}{2}+\frac{1}{2} \\ y=1 \end{gathered}[/tex]The coordinates of the circumcenter are: (-1/3,1).
In the graph it is:
