If the points $(4, - 4), (- 4, 4)$ and $(x, y)$ form an equilateral triangle, find $x$ and $y$ .

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Solution

$A B = \sqrt{64 + 64} = 8 \sqrt{2}$
$A B^{2} = B C^{2} = C A^{2}$
$\Rightarrow 128 = (x + y)^{2} + (y - 4)^{2} = (x - 4)^{2} + (y + 4)^{2}$
$∴ 128 = x^{2} + y^{2} = 8 x + 8 y + 32$
$\Rightarrow x^{2} + y^{2} + 8 x - 8 y - 96 = 0 . . . (i)$
$x = y . . . (i i)$
$x^{2} + x^{2} - 96 = 0$
$2 x^{2} = 96$
$x^{2} = 48$ $∴ x = 4 \sqrt{3}$
$C (4 \sqrt{3}, 4 \sqrt{3})$ or $C (- 4 \sqrt{3}, - 4 \sqrt{3})$

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