Find the coordinates of the point equidistant from point A(1, 2), B(3, -4)and C(5, -6).

The given three points are A(1,2) B(3,-4) and C(5,-6).
Let P (x, y) be the point equidistant from these three points.
So, PA = PB = PC

⇒ x^2 + 1– 2x + y^2 + 4 – 4y = x^2 + 9 -6x + y^2 + 16 + 8y = x 2 + 25– 10x + y^2 + 36 + 12y
⇒ – 2x– 4y + 5 = -6x + 8y +25= – 10x + 12y+61
– 2x– 4y + 5 = -6x + 8y +25
⇒ – 2x– 4y + 5 +6x – 8y -25=0
⇒ 4x– 12y -20=0
⇒ x– 3y – 5 =0….(i)
– 2x– 4y + 5 = – 10x + 12y+61
⇒- 2x– 4y + 5 +10x – 12y-61=0
⇒8x– 16y -56=0
⇒x– 2y -7=0….(ii)
Solving (i) and (ii)
x = 11, y = 2
Thus, the required point is (11, 2)

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