Question

If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that PA 2 + PB 2 = k 2 , where k is a constant.

Answer 1

The given points are A=( 3,4,5 ) and B=( −1,3,−7 ) .

Let P=( x,y,z ) be a point such that,

( PA ) 2 + ( PB ) 2 = k 2 (1)

where, k is a constant.

Distance d between two points ( x 1 , y 1 , z 1 ) and ( x 2 , y 2 , z 2 ) is,

d= ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 + ( z 2 − z 1 ) 2 (2)

Using the distance formula to find PA and PB,

PA= ( x−3 ) 2 + ( y−4 ) 2 + ( z−5 ) 2

PB= ( x− ( −1 ) 2 + ( y−3 ) 2 + ( z−( −7 ) ) 2 )

( ( x−3 ) 2 + ( y−4 ) 2 + ( z−5 ) 2 ) 2 + ( ( x− ( −1 ) 2 + ( y−3 ) 2 + ( z+7 ) 2 ) ) 2 = ( k ) 2 ( x−3 ) 2 + ( y−4 ) 2 + ( z−5 ) 2 + ( x+1 ) 2 + ( y−3 ) 2 + ( z+7 ) 2 = ( k ) 2 x 2 −6x+9+ y 2 −8y+16+ z 2 −10z+25+ x 2 +2x+1+ y 2 −6y+9+ z 2 +14z+49= k 2 2 x 2 +2 y 2 +2 z 2 −4x−14y+4z+109= k 2

Simplify further,

2 x 2 +2 y 2 +2 z 2 −4x−14y+4z+109= k 2 x 2 + y 2 + z 2 −2x−7y+2z= k 2 −109 2

Thus, the equation of theset of points P such that PA 2 + PB 2 = k 2 is given by,

x 2 + y 2 + z 2 −2x−7y+2z= k 2 −109 2 .