Show that the origin is the circum-centre of the triangle formed by the vertices (1, 0), (0, -1) and $(\frac{- 1}{2}, \frac{\sqrt{3}}{2})$ .

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Solution

$A (0, 0) B (1, 0) C (0, - 1) D (\frac{- 1}{2}, \frac{\sqrt{3}}{2})$
Let $(0, 0)$ is circumcenter, Then
$A / Q$ $A B = A C = A D$
$A B = \sqrt{1^{2}} = 1$
$A C = \sqrt{1^{2}} = 1$
$A D = \sqrt{{\frac{1}{2}}^{2} + {\frac{\sqrt{3}}{2}}^{2}} = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1$
$A B = A C = A D$
Hence $(0, 0)$ is circumcenter.

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4 : 1

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