Finding the Circumcenter of Triangle ABC

The problem asks us to find the coordinates of the circumcenter of a triangle ABC, given the equations of its sides. The circumcenter is the center of the circumscribed circle, which passes through all three vertices of the triangle. It is also the point where the perpendicular bisectors of the sides intersect.

Identifying the Triangle's Vertices

First, let's find the coordinates of the vertices A, B, and C by finding the intersection points of the given side equations:

• Side AB: $x - 2 = 0 \implies x = 2$
• Side BC: $y + 1 = 0 \implies y = -1$
• Side CA: $x + 2y - 4 = 0$

Vertex B (Intersection of AB and BC)

The intersection of $x=2$ and $y=-1$ directly gives the coordinates of Vertex B.

So, Vertex B is $(2, -1)$.

Vertex A (Intersection of AB and CA)

Substitute the equation of AB ($x=2$) into the equation of CA ($x + 2y - 4 = 0$):

$2 + 2y - 4 = 0$

$2y - 2 = 0$

$2y = 2$

$y = 1$

So, Vertex A is $(2, 1)$.

Vertex C (Intersection of BC and CA)

Substitute the equation of BC ($y=-1$) into the equation of CA ($x + 2y - 4 = 0$):

$x + 2(-1) - 4 = 0$

$x - 2 - 4 = 0$

$x - 6 = 0$

$x = 6$

So, Vertex C is $(6, -1)$.

Determining the Type of Triangle

We have the vertices A(2, 1), B(2, -1), and C(6, -1). Let's examine the nature of the sides AB and BC.

• Side AB connects (2, 1) and (2, -1). Since the x-coordinates are the same, AB is a vertical line. Its equation is $x=2$.
• Side BC connects (2, -1) and (6, -1). Since the y-coordinates are the same, BC is a horizontal line. Its equation is $y=-1$.

A vertical line and a horizontal line are perpendicular. The vertex B(2, -1) is the intersection of AB and BC. Therefore, the angle at vertex B is $90^{\circ}$.

Triangle ABC is a right-angled triangle, with the right angle at B.

Finding the Circumcenter of a Right-Angled Triangle

A key property of right-angled triangles is that their circumcenter lies exactly at the midpoint of their hypotenuse. The hypotenuse is the side opposite the right angle, which is AC in this case.

The vertices of the hypotenuse AC are A(2, 1) and C(6, -1).

Calculating the Midpoint of the Hypotenuse AC

The midpoint formula for two points $(x_1, y_1)$ and $(x_2, y_2)$ is:

Midpoint = $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

Using A(2, 1) and C(6, -1):

Circumcenter = $\left(\frac{2 + 6}{2}, \frac{1 + (-1)}{2}\right)$

Circumcenter = $\left(\frac{8}{2}, \frac{0}{2}\right)$

Circumcenter = $(4, 0)$

The coordinates of the circumcenter of triangle ABC are (4, 0).

Verification

To verify, we can check if the distance from (4, 0) to each vertex is the same (this distance is the circumradius).

• Distance from (4, 0) to A(2, 1): $\sqrt{(4-2)^2 + (0-1)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4+1} = \sqrt{5}$
• Distance from (4, 0) to B(2, -1): $\sqrt{(4-2)^2 + (0-(-1))^2} = \sqrt{2^2 + 1^2} = \sqrt{4+1} = \sqrt{5}$
• Distance from (4, 0) to C(6, -1): $\sqrt{(4-6)^2 + (0-(-1))^2} = \sqrt{(-2)^2 + 1^2} = \sqrt{4+1} = \sqrt{5}$

Since all distances are equal, (4, 0) is indeed the circumcenter.

Revision Table: Triangle and Circumcenter

Additional Information on Circumcenter and Triangles

The circumcenter is a fundamental point in triangle geometry. Its location depends on the type of triangle:

• Acute Triangle: The circumcenter lies inside the triangle.
• Right Triangle: The circumcenter lies on the midpoint of the hypotenuse.
• Obtuse Triangle: The circumcenter lies outside the triangle.

The distance from the circumcenter to any vertex is called the circumradius. The perpendicular bisectors of the sides are lines perpendicular to each side, passing through the midpoint of that side. Their intersection point is always the circumcenter.

For a general triangle (not necessarily right-angled), finding the circumcenter involves finding the equations of at least two perpendicular bisectors and solving them simultaneously. However, recognizing the right angle in this problem significantly simplified the process, allowing us to use the hypotenuse midpoint property.