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Solution

In equilateral triangle AB = BC = AC
$AC= \sqrt{\left ( x+4 \right )^{2}+\left ( y-3 \right )^{2}}$
$AC= \sqrt{x^{2}+16+8x+y^{2}+9-6y}$
$\left ( because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\; \;\left ( a-b \right )^{2} = a^{2}+b^{2}-2ab \right )$
$BC= \sqrt{\left ( x-4 \right )^{2}+\left ( y-3 \right )^{2}}$
$BC= \sqrt{x^{2}+16-8x+y^{2}+9-6y}$      $\left ( because \left ( a-b \right )^{2} = a^{2}+b^{2}-2ab \right )$
AC = BC
$\sqrt{x^{2}+16+8x+y^{2}+9-6y}= \sqrt{x^{2}+16-8x+y^{2}+9-6y}$

Squaring both side
8x + 8x = 0
16x = 0
x = 0
C = (0, y)
$Length\, of\, AB= \sqrt{\left ( 4+4 \right )^{2}+\left ( 3-3 \right )^{2}}$
$AB= \sqrt{\left ( 8 \right )^{2}}= 8$
AC = AB
$\sqrt{x^{2}+16+8x+y^{2}+9-6y}= 8$

Put x = 0, squaring both side
0 + 16 + 0 + y2 + 9 – 6y = 64
y2 – 6y + 25 – 64 = 0
y2 – 6y – 39 = 0
$y= \frac{6\pm \sqrt{36+156}}{2}$
$y=\frac{6-8\sqrt{3}}{2}$       (For origin in the interior we take the only term with negative sign)
$y=3-4\sqrt{3}$

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