# Find the equation of the set of the points P such that its distances from the points A(3, 4, -5) and B(-2, 1, 4) are equal.

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Find the equation of the set of the points P such that its distances from the points A(3, 4, -5) and B(-2, 1, 4) are equal.

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Given: Points are A(3, 4, -5) and B(-2, 1, 4)

To find: the equation of the set of the points, i.e. locus of points which are equidistant from the given points

Let the required point P(x, y, z)

According to the question:

PA = PB

⇒ PA2 = PB2

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

$\sqrt{(a−m)^2+(b−n)^2+(c−o)^2}$

Therefore,

The distance between P(x, y, z) and A(3, 4, -5) is PA,

=$\sqrt{(x-3)^2+(y−4)^2+(z-(-5))^2}$

=$\sqrt{(x-3)^2+(y-4)^2+(z+5)^2}$

The distance between P(x, y, z) and B(-2, 1, 4) is PB,

=$\sqrt{(x-(-2))^2+(y−1)^2+(z-4)^2}$

$\sqrt{(x+2)^2+(y−1)^2+(z-4)^2}$

As PA2 = PB2

(x – 3)2+ (y – 4)2 + (z + 5)2 = (x + 2)2 + (y – 1)2 + (z – 4)2

⇒ x2+ 9 – 6x + y2 + 16 – 8y + z2 + 25 + 10z = x2+ 4 + 4x + y2 + 1 – 2y + z2 + 16 – 8z

⇒ x2+ 9 – 6x + y2 + 16 – 8y + z2 + 25 + 10z – x2– 4 – 4x – y2 – 1 + 2y – z2 – 16 + 8z = 0

⇒ – 6x – 6y + 18z + 29 = 0

⇒ 6x + 6y – 18z – 29 = 0

Hence locus of point P is 6x + 6y – 18z – 29 = 0