# Find the locus of P if PA2 + PB2 = 2k2, where A and B are the points (3, 4, 5) and (-1, 3, -7).

Given: Points are A(3, 4, 5) and B(-1, 3, -7)

To find: the locus of point P which moves in such a way that PA2 + PB2 = 2k2

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

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(3, 4, 5) is PA, Distance between P(x, y, z) and B(-1, 3, -7) is PB,  According to question:

PA2 + PB2 = 2k2

(x – 3)2+ (y – 4)2 + (z – 5)2 + (x + 1)2 + (y – 3)2 + (z + 7)2 = 2k2

x2+ 9 – 6x + y2 + 16 – 8y + z2 + 25 – 10z + x2+ 1 + 2x + y2 + 9 – 6y + z2 + 49 + 14z = 2k2

2x2+ 2y2 + 2z2 – 4x – 14y + 4z + 109 = 2k2

2x2+ 2y2 + 2z2 – 4x – 14y + 4z + 109 – 2k2 = 0

Hence locus of point P is 2x2+ 2y2 + 2z2 – 4x – 14y + 4z + 109 – 2k2 = 0

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