On the line segment joining $(1, 0)$ and $(3, 0)$ , an equilateral triangle is drawn having its vertex in the fourth quadrant, then radical centre of the circles described on its sides as diameter is

(3, - \frac{1}{\sqrt{3}})

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(3, - \sqrt{3})

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(2, - \frac{1}{\sqrt{3}})

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(2, - \sqrt{3})

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Solution

The correct option is C $(2, - \frac{1}{\sqrt{3}})$
Radical centre of the cirlces described on the sides of a triangle as diameters is the orthocentre of the triangle.
$∴$ $D = (2, 0)$
$D H = - B C tan \frac{π}{6} = - \frac{1}{\sqrt{3}}$
$∴$ Coordinates of H are $(2, - \frac{1}{\sqrt{3}})$

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On the line segment joining (1,0) and (3,0), an equilateral triangle is drawn having its vertex in the fourth quadrant, then radical centre of the circles described on its sides as diameter is

The correct option is C (2,−1√3)Radical centre of the cirlces described on the sides of a triangle as diameters is the orthocentre of the triangle.∴ D=(2,0) DH=−BCtanπ6=−1√3∴ Coordinates of H are (2,−1√3)

On the line segment joining $(1, 0)$ and $(3, 0)$ , an equilateral triangle is drawn having its vertex in the fourth quadrant, then radical centre of the circles described on its sides as diameter is

The correct option is C $(2, - \frac{1}{\sqrt{3}})$
Radical centre of the cirlces described on the sides of a triangle as diameters is the orthocentre of the triangle.
$∴$ $D = (2, 0)$
$D H = - B C tan \frac{π}{6} = - \frac{1}{\sqrt{3}}$
$∴$ Coordinates of H are $(2, - \frac{1}{\sqrt{3}})$