If two vertices of an equilateral triangle ar | vedclass

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Question

(A) Let the vertices of the equilateral triangle be $A(-1, 0)$,$B(1, 0)$,and $C(0, y_c)$.Since the triangle is equilateral,the distance $AB = BC = AC$

Vedclass · Accepted Answer

$x^2 + (y - \frac{1}{\sqrt{3}})^2 = \frac{4}{3}$

Vedclass · Answer

(A) Let the vertices of the equilateral triangle be $A(-1, 0)$,$B(1, 0)$,and $C(0, y_c)$.Since the triangle is equilateral,the distance $AB = BC = AC$.The length of side $AB = \sqrt{(1 - (-1))^2 + (0 - 0)^2} = 2$.The height of the triangle is $h = \frac{\sqrt{3}}{2} 	imes 	ext{side} = \frac{\sqrt{3}}{2} 	imes 2 = \sqrt{3}$.The midpoint of $AB$ is $(0, 0)$. The third vertex $C$ lies on the perpendicular bisector of $AB$,which is the $y$-axis. Thus,$C = (0, \sqrt{3})$ or $(0, -\sqrt{3})$.The circumcenter of an equilateral triangle is its centroid. For $C = (0, \sqrt{3})$,the centroid is $(\frac{-1+1+0}{3}, \frac{0+0+\sqrt{3}}{3}) = (0, \frac{1}{\sqrt{3}})$.The radius $R$ is the distance from the centroid $(0, \frac{1}{\sqrt{3}})$ to $(1, 0)$,so $R^2 = (1-0)^2 + (0 - \frac{1}{\sqrt{3}})^2 = 1 + \frac{1}{3} = \frac{4}{3}$.The equation of the circumcircle is $x^2 + (y - \frac{1}{\sqrt{3}})^2 = \frac{4}{3}$.Similarly,for $C = (0, -\sqrt{3})$,the centroid is $(0, -\frac{1}{\sqrt{3}})$,giving $x^2 + (y + \frac{1}{\sqrt{3}})^2 = \frac{4}{3}$.

If two vertices of an equilateral triangle are $(-1, 0)$ and $(1, 0)$,then its circumcircle is:

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