Find the third vertices of equilateral triangle whose vertices are (3,4) and (-2,3)

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Solution

Let the side is the equilateral triangle be A(x,y), B(3,4) and C(-2, 3)
For the triangle to be equilateral, all AB= BC= AC
Using, AB= AC
$(x - 3)^{2} + (y - 4)^{2} = (x + 2)^{2} + (y - 3)^{2} x^{2} - 6 x + 9 + y^{2} - 8 y + 16 = x^{2} + 4 x + 4 + y^{2} - 6 y + 9 12 = 10 x + 2 y 6 = 5 x + y$ ...(1)
Using AB = BC
$(x - 3)^{2} + (y - 4)^{2} = (3 + 2)^{2} + (4 - 3)^{2} (x - 3)^{2} + (6 - 5 x - 4)^{2} = 26 26 x^{2} - 26 x - 13 = 0$
On solving
$x = \frac{1 + \sqrt{3}}{2} o r \frac{1 - \sqrt{3}}{2} y = \frac{7 - 5 \sqrt{3}}{2} o r y = \frac{7 + 5 \sqrt{3}}{2}$

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