Question

Show that (2, 1) is the centre of the circumcircle of the triangle whose vertices are (-3, -9), (13, -1) and (-9,3)​

Answer 1

Answer:

$hence \: \: \: p(2 \: \: \: \: 1) \: is \: circumcenter \: of \: triangle \: abc.$

Step-by-step explanation:

$let \: \: \: \: \: \: a( - 3 \: \: \: \: - 9) \: \: \: \: \: \: \: b(13 \: \: \: \: - 1) \: \: \: \: \: \: \: \: c( - 9 \: \: \: \: 3) \: \: \: and \: \: \: \: p(2 \: \: \: \: 1) \\ \\ then \: \: \: \: ap = \sqrt{ {(2 + 3)}^{2} + {(1 + 9)}^{2} } \\ \\ = \sqrt{25 + 100} \\ \\ = \sqrt{125} \\ \\ \\ bp = \sqrt{ {(13 - 2)}^{2} + {(1 + 1)}^{2} } \\ \\ = \sqrt{121 + 4} \\ \\ = \sqrt{125} \\ \\ \\ cp = \sqrt{ {(2 + 9)}^{2} + {(3 - 1)}^{2} } \\ \\ = \sqrt{125} = \\ \\ therefore \: \: \: ap = bp = cp = \sqrt{125}=RADIUS. \\ \\ p \: is \: equidistant \: from \: a \: \: b \: \: \: and \: \: \: c. \\ \\ hence \: \: \: p(2 \: \: \: \: 1) \: is \: circumcenter \: of \: triangle \: abc.$