QQuestionMathematics
QuestionMathematics
Find the coordinates of the circumcenter of the triangle with the given vertices:
A(2, 6), B(8, 6), C(8, 10)
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Answer
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Step 1:I'll solve this step-by-step using precise LaTeX formatting:
Step 2:: Understand the Circumcenter
The circumcenter is the point where the perpendicular bisectors of the triangle's sides intersect. This point is equidistant from all three vertices and is the center of the circumscribed circle.
Step 3:: Calculate the Midpoints
- Midpoint of BC: $$\left(\frac{8+8}{2}, \frac{6+10}{2}\right) = (8, 8)
First, let's find the midpoints of two sides:
Step 4:: Find Slopes of Original Sides
- Slope of AB: $$m_{AB} = \frac{6-6}{8-2} = 0
- Slope of BC: m_{BC} = \frac{10 - 6}{8 - 8} = \text{undefined}
Step 5:: Calculate Perpendicular Bisector Slopes
- Perpendicular slope to BC: $$0
- Perpendicular slope to AB: undefined (vertical line)
Step 6:: Write Perpendicular Bisector Equations
- First perpendicular bisector (through midpoint of AB): $$x = 5
- Second perpendicular bisector (through midpoint of BC): y = 8
Step 7:: Find Intersection Point
The intersection of $$x = 5$$ and $$y = 8$$ gives the circumcenter:
\text{Circumcenter} = (5, 8)
Final Answer
Verification: - Distance from (5, 8) to A(2, 6): \sqrt{(5 - 2)^{2} + (8 - 6)^{2}} = \sqrt{3^{2} + 2^{2}} = \sqrt{9 + 4} = \sqrt{13} - Distance from (5, 8) to B(8, 6): \sqrt{(5 - 8)^{2} + (8 - 6)^{2}} = \sqrt{(- 3)^{2} + 2^{2}} = \sqrt{9 + 4} = \sqrt{13} - Distance from (5, 8) to C(8, 10): \sqrt{(5 - 8)^{2} + (8 - 10)^{2}} = \sqrt{(- 3)^{2} + (- 2)^{2}} = \sqrt{9 + 4} = \sqrt{13}
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