Find the locus of P if PA2 + PB2 = 2k2 where A and B are the points (3,4,5) and (-1,3,-7).
Asked by Topperlearning User
| 20th Feb, 2015,
10:30: AM
Expert Answer:
Let P(x,y,z) be the point whose locus is to be found, Now
PA2 + PB2 = 2k2.
(x - 3)2 + (y - 4)2 + (z - 5)2 + (x + 1)2 + (y - 3)2 + (z + 7)2 =2k2
x2 - 6x + 9 + y2 - 8y + 16 + z2 - 10z + 25 + x2 + 2x + 1 + y2 - 6y + 9 + z2 + 14z + 49 = 2k2
2x2 + 2y2 + 2z2 - 4x - 14y + 4z + 109 = 2k2.
Answered by
| 20th Feb, 2015,
12:30: PM
Concept Videos
- Find the distance between the points P and Q having coordinates (2,3,1) and (-2,4,-1).
- Using distance formula prove that the points A(4,-3,-1), B(5,-7,6) and C(3,1,-8) are collinear.
- Determine the point in XY - plane which is equidistant from three points A(2,0,3) B(0,3,2) and C(0,0,1).
- Verify that the given points are the vertices of an isosceles triangle, A(0,7,-10), B(1,6,-6) and C(4,9,-,6).
- Prove that the points A(1,3,0), B(-5,5,2) C(-9,-1,2) and D(-3,-3,0) are the vertices of a parallelogram. Also show that ABCD is not a rectangle.
- If A (-2,2,3) and B(13,-3,13) are two points such that 3PA = 2PB, find the locus of P.
- Show that A(a,b,c), B(b,c,a) and C(c,a,b) are the vertices of an equilateral triagle.
- Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (-4, 0, 0) is equal to 10?
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change