Answer:
(-7,6)
Step-by-step explanation:
The vertices of a triangle ABC are A(-8,5) B(-6,7) C(-8,7).
Distance formula:
[tex]D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Using distance formula we get
[tex]a=BC=\sqrt{\left(-8-\left(-6\right)\right)^2+\left(7-7\right)^2}=2[/tex]
[tex]b=AC=\sqrt{\left(-8-\left(-8\right)\right)^2+\left(7-5\right)^2}=2[/tex]
[tex]c=AB=\sqrt{\left(-6-\left(-8\right)\right)^2+\left(7-5\right)^2}=2\sqrt{2}[/tex]
Using these sides, we get
[tex]AB^2=(2\sqrt{2})^2=8[/tex]
[tex]BC^2+AC^2=2^2+2^2=8[/tex]
Since [tex]AB^2=BC^2+AC^2[/tex], therefore triangle ABC is a right angled triangle.
Circumcenter of a right angled triangle is the midpoint of hypotenuse.
In triangle ABC, side AB is hypotenuse. So, circumcenter of the triangle ABC is the midpoint of AB.
[tex]Circumcenter=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})[/tex]
[tex]Circumcenter=(\frac{-8-6}{2}, \frac{5+7}{2})[/tex]
[tex]Circumcenter=(\frac{-14}{2}, \frac{12}{2})[/tex]
[tex]Circumcenter=(-7,6)[/tex]
Therefore, the circumcenter of the triangle ABC is (-7,6).
