Q22 of 84 Page 114

Show that (2, 1) is the circum–center of the triangle formed by the vertices (3, 1), (2, 2) and (1, 1).

Formula used:


Let the points be A (3, 1), B (2, 2), C (1, 1) and S(2, 1)


Distance of SA


⇒ SA = √ ((3 – 2)2 + (1 – 1)2)


⇒ SA = √ ((1)2 + (0)2


⇒ SA = √ (1 + 0)


⇒ SA = √ 1 = 1


Distance of SB


⇒ SB = √ ((2 – 2)2 + (2 – 1)2)


⇒ SB = √ ((0)2 + (1)2


⇒ SB = √ (0 + 1)


⇒ SB = √ 1 = 1


Distance of SC


⇒ SC = √ ((1 – 2)2 + (1 – 1)2)


⇒ SC = √ ((–1)2 + (0)2


⇒ SC = √ (1 + 0)


⇒ 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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