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Answer:
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Step-by-step explanation:
Coordinates of circumcenter will be [tex](-\frac{1}{3},1)[/tex].
Property of a circumcenter,
"Distances of all vertices from circumcenter is same"
If A(h, k) is the circumcenter of the triangle PQR,
PA² = QA² = RA²
By using the expression for the distance between two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex],
Distance = [tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
For PA² = QA²,
[tex](\sqrt{(-2-h)^2+(5-k)^2})^2=(\sqrt{(4-h)^2+(k-1)^2})^2[/tex]
4 + h² + 4h + 25 + k² - 10k = 16 + h² - 8h + k² + 1 - 2k
12h - 8k = -12
3h - 2k = -3 --------(1)
For QA² = RA²,
[tex](\sqrt{(4-h)^2+(1-k)^2})^2=(\sqrt{(-2-h)^2+(-3-k)^2})^2[/tex]
16 + h² - 8h + k² + 1 - 2k = 4 + h² + 4h + k² + 9 + 6k
-12h - 8k = -4
12h + 8k = 4
3h + 2k = 1 ------ (2)
Add equation (1) and (2),
3h - 2k + 3h + 2k = -3 + 1
6h = -2
h = [tex]-\frac{1}{3}[/tex]
From equation (1)
[tex]3(-\frac{1}{3})-2k=-3[/tex]
-2k = -2
k = 1
Therefore, coordinates of circumcenter will be [tex](-\frac{1}{3},1)[/tex].
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