Question

What are the coordinates of the circumcenter of this triangle? Enter your answer in the boxes.

Answer 1

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: $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

AB=$\sqrt{(-3-5)^2+(-1+1)^2}=8$ units

BC=$\sqrt{(5+1)^2+(-3+3)^2}=6$ units

AC=$\sqrt{(5+1)^2+(-3-5)^2}=10$ units

Pythagoras theorem:$(Hypotenuse)^2=(Base)^2+(Perpendicular\;side)^2}$

$AB^2+BC^2=8^2+6^2=100$

$AC^2=(10)^2=100$

Therefore, $AC^2=AB^2+BC^2$

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:$x=\frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2}$

Using this formula then, we get

Mid-point of hypotenuse AC is given by

$x=\frac{-1+5}{2}=2,y=\frac{5-3}{2}=1$

Hence, the circum-center of triangle is (2,1).