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If the point D $(x, y)$ is equidistant from the points E $(0, 3)$ and F $(3, 0)$, prove that $x = y$. Hence, find the $x$ coordinate of the point D, if $\triangle DEF$ is an equilateral triangle.
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Ans. $\sqrt{(x-0)^2 + (y-3)^2} = \sqrt{(x - 3)^2 + (y - 0)^2}$ (1 Mark)
$x^2 + y^2 - 6y + 9 = x^2 - 6x + 9 + y^2$ (1/2 Mark)
$x=y$ (1/2 Mark)
As $\triangle DEF$ is an equilateral triangle
$\therefore x^2 + y^2 - 6y + 9 = 18$ (using $DE = DF$) (1 Mark)
$2x^2-6x-9=0$ (using $x = y$) (1/2 Mark)
Solving we get
$x = \frac{6 \pm 6\sqrt{3}}{4}$ or $\frac{3 \pm 3\sqrt{3}}{2}$ (1/2 Mark)
$x^2 + y^2 - 6y + 9 = x^2 - 6x + 9 + y^2$ (1/2 Mark)
$x=y$ (1/2 Mark)
As $\triangle DEF$ is an equilateral triangle
$\therefore x^2 + y^2 - 6y + 9 = 18$ (using $DE = DF$) (1 Mark)
$2x^2-6x-9=0$ (using $x = y$) (1/2 Mark)
Solving we get
$x = \frac{6 \pm 6\sqrt{3}}{4}$ or $\frac{3 \pm 3\sqrt{3}}{2}$ (1/2 Mark)