Answer: They're equal to each other yes corresponding parts of congruent triangles are equal
Step-by-step explanation:
Answer:
Step-by-step explanation:
<u>Given vertices of a triangle:</u>
- 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.
<u>Use distance formula:</u>
- AM = √(x - 4)² + (y - 12)²
- BM = √(x - 14)² + (y - 6)²
- CM = √(x + 6)² + (y - 2)²
<u>All three distances are equal:</u>
- AM = BM = CM ⇒ AM² = BM² = CM² ⇒
- (x - 4)² + (y - 12)² = (x - 14)² + (y - 6)² = (x + 6)² + (y - 2)²
<u>Square of each:</u>
- 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
<u>Comparing the squares, we get following system:</u>
- -8x- 24y + 160 = -28x - 12y + 232 ⇒ 20x - 12y = 72 ⇒ 5x - 3y = 18
- - 8x - 24y + 160 = 12x - 4y + 40 ⇒ 20x + 20y = 120 ⇒ x + y = 6
<u>Solve by substitution, x = 6 - y:</u>
- 5(6 - y) - 3y = 18
- 30 - 8y = 18
- 8y = 12
- y = 1.5
<u>Find x:</u>
So the circumcenter is M(4.5, 1.5)
