Question

Show that the point (1,-1) is the center of the circular circumscribing the triangle whose vertices are (4,3), (-2,3) and (6,-1)

Answer 1

To show that point $0(1,-1)$ is a circumcentre of triangle formed by vertices say $A(4,3),B(-2,3),C(6,-1)$

We need to show $OA=OB=OC$

$OA=\sqrt{(4-1{)}^2+(3+1{)}^2}=\sqrt{3^2+4^2}=5$

$OB=\sqrt{(-2-1{)}^2+(3+1{)}^2}=\sqrt{3^2+4^2}=5$

$OC=\sqrt{(6-1{)}^2+(-1+1{)}^2}=\sqrt{5^2+0^2}=5$

Since $OA=OB=OC$

O is circumcentre of triangle ABC.