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If the point A $(x, y)$ is equidistant from the points B $(-2, 0)$ and C $(2, 0)$,
prove that the point A lies on y-axis. Also, find the coordinates of the point
A, if $\triangle ABC$ is an equilateral triangle.
prove that the point A lies on y-axis. Also, find the coordinates of the point
A, if $\triangle ABC$ is an equilateral triangle.
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Ans. $\sqrt{(x + 2)^2 + (y - 0)^2} = \sqrt{(x - 2)^2 + (y - 0)^2}$ (1 Mark)
getting $x = 0$ (1/2 Mark)
As $x = 0$ so A lies on y axis (1/2 Mark)
As triangle is equilateral, $AB = BC = AC$ gives
$\sqrt{(0 + 2)^2 + (y - 0)^2} = \sqrt{(-2 - 2)^2 + (0 - 0)^2}$ (1/2 Mark)
which gives $y^2 = 12$ (1/2 Mark)
Thus, coordinates of point A are $(0, \pm 2\sqrt{3})$ (1 Mark)
getting $x = 0$ (1/2 Mark)
As $x = 0$ so A lies on y axis (1/2 Mark)
As triangle is equilateral, $AB = BC = AC$ gives
$\sqrt{(0 + 2)^2 + (y - 0)^2} = \sqrt{(-2 - 2)^2 + (0 - 0)^2}$ (1/2 Mark)
which gives $y^2 = 12$ (1/2 Mark)
Thus, coordinates of point A are $(0, \pm 2\sqrt{3})$ (1 Mark)