Answer :

Formula used:


Let the points be A (1, 0), B (0, –1), C and S (0, 0)


Distance of SA


SA = √ ((1 – 0)2 + (0 – 0)2)


SA = √ ((1)2 + (0)2


SA = √ (1 + 0)


SA = √ 1 = 1


Distance of SB


SB = √ ((0 – 0)2 + (–1 – 0)2)


SB = √ ((0)2 + (–1)2


SB = √ (0 + 1)


SB = √ 1 = 1


Distance of SC


SC


SC


SC


SC


SC =√ 1 = 1


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 the circum–centre of the triangle ABC.


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