Find the coordinates of the point equidistant from three given points A(5,1), B(-3.-7) and
C(7,-1).
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The given three points are A (5, 1) , B (–3, –7 ) and C (7, –1)
Let P (x, y) be the point equidistant from these three points.
So, PA = PB = PC
⇒ x² + 25 – 10x + y² + 1 – 2y = x² + 9 + 6x + y² + 49 + 14y = x² + 49 – 14x + y² + 1 + 2y
⇒ 25 – 10x + 1 – 2y = 9 + 6x + 49 + 14y = 49 – 14x + 1 + 2y
25 – 10x + 1 – 2y = 9 + 6x + 49 + 14y
⇒ 26 = 16x + 58 + 16y
⇒ 16x + 32 + 16y = 0
⇒ x + 2 + y = 0 … (1)
25 – 10x + 1 – 2y = 49 – 14x + 1 + 2y
⇒ 26 + 4x – 4y = 50
⇒ 4x – 4y = 24
⇒ x – y = 6 … (2)
Solving (1) and (2)
x = 2, y = –4
Thus, the required point is (2, –4)
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