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

CONCEPT:

The distance between the points A(x1, y1, z1) and B(x2, y2, z2) is given by:

\(d = \sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\)

CALCULATION:

Let P(x, y, z) be the point which is equidistant from the points A(3, 4, -5) and B(-2, 1, 4)

As we know that, the distance between the points A(x1, y1, z1) and B(x2, y2, z2) is given by:

\(d = \sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\)

First let's find out the distance between the the points P and A

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

Similarly, let's find out the distance between the the points P and B

⇒ \(PB = \sqrt {(-2- x)^2 + (1 - y)^2 + (4 - z)^2}\)

∵ PA = PB

⇒ \(\sqrt {(3- x)^2 + (4 - y)^2 + (-5 - z)^2} = \sqrt {(-2- x)^2 + (1 - y)^2 + (4 - z)^2}\)

By squaring both the sides we get,

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

⇒ x2 + y2 + z2 - 6x - 8y + 10z + 50 = x2 + y2 + z2 + 4x - 2y - 8z + 21

⇒ 10x + 6y - 18z - 29 = 0

So, the set of the points which are equidistant from the points A(3, 4, -5) and B(-2, 1, 4) is given by: 10x + 6y - 18z - 29 = 0.

Hence, correct option is 2.