Question

Prove that the point $(-\frac{1}{14},\frac{39}{14})$ is the centre of the circle circumscribing the triangle whose angular points are (1, 1), (2, 3), and ( - 2, 2).

Answer 1

Given that points of triangle $(1,1),(2,3),(-2,2)$

Given circumcentre $(\frac{-1}{14},\frac{39}{14})$
By distance formula we can check the point as circumcentre
$\sqrt{{(1+\frac{1}{14})}^2+{(1-\frac{39}{14})}^2}=\sqrt{{(2+\frac{1}{14})}^2+{(3-\frac{39}{14})}^2}=\sqrt{{(-2+\frac{1}{14})}^2+{(2-\frac{39}{14})}^2}=\frac{\sqrt{850}}{14}$
Then, circumradius $R=\frac{\sqrt{850}}{14}$
Hence proved