Q. 235.0( 1 Vote )

# Show that the origin is the circum–center of the triangle formed by the vertices (1, 0), (0, −1) and .

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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