Question

20 pts

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Answer 1

Let the vertices be A (5,1) , B(11,1) and C(11,9) .

Slope of AB = (1-1)/(11-5) = 0 (parallel to X-axis)

Slope of BC = (9-1)/(11-11) = ∞ (parallel to Y-axis)

So AB⊥BC with ∠B = 90° ⇒  ΔABC is right triangle.

⇒ Circum center is midpoint of  hypotenuse (i.e AC)

So circumcenter =  ((5+11)/2 , (9+1)/2) = (8,5)

Answer 2

Let the vertices be A (5,1) , B(11,1) and C(11,9) .

Slope of AB = (1-1)/(11-5) = 0 (parallel to X-axis)

Slope of BC = (9-1)/(11-11) = ∞ (parallel to Y-axis)

So AB⊥BC with ∠B = 90° ⇒  ΔABC is right triangle.

⇒ Circum center is midpoint of  hypotenuse (i.e AC)

So circumcenter =  ((5+11)/2 , (9+1)/2) = (8,5)

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