Question

Mhanifa please help! Find the coordinates of the circumcenter of the triangle with the vertices A(4, 12), B(14, 6), and C(-6, 2).

Answer 1

• (4.5, 1.5)

Explanation:

Given vertices of a triangle:

• A(4, 12), B(14, 6), and C(-6, 2)

Let the circumcenter is point M(x, y)

We know that the circumcenter is equidistant from each of the vertices.

Use distance formula:

• AM = √(x - 4)² + (y - 12)²
• BM = √(x - 14)² + (y - 6)²
• CM = √(x + 6)² + (y - 2)²

All three distances are equal:

• AM = BM = CM ⇒ AM² = BM² = CM² ⇒
• (x - 4)² + (y - 12)² = (x - 14)² + (y - 6)² = (x + 6)² + (y - 2)²

Square of each:

• AM² = x² - 8x + 16 + y² - 24y + 144 = x² + y² - 8x - 24y + 160
• BM² = x² - 28x + 196 + y² - 12y + 36 = x² + y² - 28x - 12y + 232
• CM² = x² + 12x + 36 + y² - 4y + 4 = x² + y² + 12x - 4y + 40

Comparing the squares, we get following system:

• -8x- 24y + 160 = -28x - 12y + 232 ⇒ 20x - 12y = 72 ⇒ 5x - 3y = 18
• - 8x - 24y + 160 = 12x - 4y + 40 ⇒ 20x + 20y = 120 ⇒ x + y = 6

Solve by substitution, x = 6 - y:

• 5(6 - y) - 3y = 18
• 30 - 8y = 18
• 8y = 12
• y = 1.5

Find x:

• x = 6 - 1.5 = 4.5

So the circumcenter is M(4.5, 1.5)