Answer:
(2,1)
Step-by-step explanation:
We can see that the given triangle.
The coordinates of A are (-1,5) .
The coordinates of B are (-1,-3).
The coordinates of C are (5,-3).
Distance formula: [tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
AB=[tex]\sqrt{(-3-5)^2+(-1+1)^2}=8[/tex] units
BC=[tex]\sqrt{(5+1)^2+(-3+3)^2}=6[/tex] units
AC=[tex]\sqrt{(5+1)^2+(-3-5)^2}=10[/tex] units
Pythagoras theorem:[tex](Hypotenuse)^2=(Base)^2+(Perpendicular\;side)^2}[/tex]
[tex]AB^2+BC^2=8^2+6^2=100[/tex]
[tex]AC^2=(10)^2=100[/tex]
Therefore, [tex]AC^2=AB^2+BC^2[/tex]
When a triangle satisfied the Pythagoras theorem then, the triangle is right triangle.
Hence, the given triangle is a right triangle.
We know that circum-center of right triangle is the mid point of hypotenuse.
Mid-point formula:[tex]x=\frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2}[/tex]
Using this formula then, we get
Mid-point of hypotenuse AC is given by
[tex]x=\frac{-1+5}{2}=2,y=\frac{5-3}{2}=1[/tex]
Hence, the circum-center of triangle is (2,1).
