Question

show that point(1,-1) is the centre of circle circumscribing in triangle whose vertices are (4,3),(-2,3) and (6,-1)

Answer 1

let the triangle be PQR.
with p(4,3), q(-2,3), and r(6,-1)
and the given point be o(1,-1)
then you can see the solution in the following picture

Answer 2

Answer:

$(1,-1)\:is \: circumcentre \\ of \: \triangle ABC$

Step-by-step explanation:

$Let \:A(4,3),\:B(-2,3)\:and \\ C(6,-1) \: are \: vertices \: of \\ a \: triangle.$

$Let \: P(1,-1) \: is \: a \\ interior\: point \: in \: the \\ circle$

$We\: know \: the \\\boxed {distance \: formula \:d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}$

$Distance \: PA = \sqrt{(4-1)^{2}+(3+1)^{2}}\\=\sqrt{3^{2}+4^{2}}\\=\sqrt{9+16}\\=\sqrt{25}\\=5$

$Distance \: PB = \sqrt{(-2-1)^{2}+(3+1)^{2}}\\=\sqrt{3^{2}+4^{2}}\\=\sqrt{9+16}\\=\sqrt{25}\\=5$

$Distance \: PC= \sqrt{(6-1)^{2}+(-1+1)^{2}}\\=\sqrt{5^{2}+0^{2}}\\=\sqrt{25}\\=5$

$Now, PA = PB = PC = 5$

$P\: is \: equidistant \\ from \: three \: vertices \: of \: \triangle ABC$

$P \: circumcentre \: of \\ \triangle ABC$

Therefore,.

$(1,-1)\:is \: circumcentre \\ of \: \triangle ABC$

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