Q. 215.0( 1 Vote )

# Find the locus of the points which are equidistant from the points (1, 2, 3) and (3, 2, -1).

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

To find: the 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,

Therefore,

The distance between P(x, y, z) and A(1, 2, 3) is PA,

The distance between P(x, y, z) and B(3, 2, -1) is PB,

As PA2 = PB2

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

x2+ 1 – 2x + y2 + 4 – 4y + z2 + 9 – 6z = x2+ 9 – 6x + y2 + 4 – 4y + z2 + 1 + 2z

x2+ 1 – 2x + y2 + 4 – 4y + z2 + 9 – 6z – x2– 9 + 6x – y2 – 4 + 4y – z2 – 1 – 2z = 0

4x – 8z = 0

4(x – 2z) = 0

x – 2z = 0

Hence locus of point P is x – 2z = 0

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Understand the Concept of Locus57 mins
Lecture on radical axis of 2 circles48 mins
Parametric Equations of Straight line48 mins
Interactive Quiz on Common tangents & angle of intersection of 2 circles56 mins
Standard Equation of Circle54 mins
Interactive Quiz on Centroid, incentre, orthocentre & circumcentre56 mins
Various Forms of Equations of line45 mins
Lecture on Common tangents & angle of intersection of 2 circles56 mins
Properties of tangents to parabola42 mins
Lecture on Tangents to a Circle57 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses