Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
2.3k views
in 3D Coordinate Geometry by (32.9k points)
closed by

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.

1 Answer

+1 vote
by (30.7k points)
selected by
 
Best answer

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

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...