Answer: The coordinates of circumcenter is (1,3).
Explanation:
It is given that the triangle have vertices A(0,1), B(2, 1) , and C(2, 5).
The distance formula,
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex]AB=\sqrt{(2-0)^2+(1-1)^2}=2[/tex]
[tex]BC=\sqrt{(2-2)^2+(5-1)^2}=4[/tex]
[tex]AC=\sqrt{(2-0)^2+(5-1)^2}=\sqrt{20}[/tex]
Since,
[tex](AC)^2=(AB)^2+(BC)^2[/tex]
By pythagoras we can say that the given triangle is a right angle triangle and AC is the hypotenuse of the triangle.
The circumcentre of a right angle triangle is the midpoint of the hypotenuse.
Midpoint of AC,
[tex]\text{Midpoint of AC}=(\frac{0+2}{2}, \frac{1+5}{2})[/tex]
[tex]\text{Midpoint of AC}=(\frac{2}{2}, \frac{6}{2})[/tex]
[tex]\text{Midpoint of AC}=(1,3)[/tex]
Therefore, the coordinates of circumcenter is (1,3).
