Question

Show that $S(4,3)$ is the circum-centre of the triangle joining the points $A(9,3),B(7,-1)$ and$C(1,-1)$.

Answer 1

$SA=\sqrt{(9-4{)}^2+(3-3{)}^2}=\sqrt{25}=5$
$SB=\sqrt{(7-4{)}^2+(-1-3{)}^2}=\sqrt{25}=5$
$SC=\sqrt{(1-4{)}^2+(-1-3{)}^2}=\sqrt{25}=5$
$\therefore SA=SB=SC$.
It is known that the circum-centre is equidistant from all the vertices of a triangle.
Since, S is equidistant from all the three vertices, it is thee circum-centre of the triangle ABC.