Question

Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

Answer 1

The given points are A=( 1,2,3 ) and B=( 3,2,−1 ) .

Let P=( x,y,z ) be a point which is equidistant from both A and B.

The formula to find the 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 

To find the distance AP between the points A and P, the value of x 1 is 1 , x 2 is x , y 1 is 2 , y 2 is y , z 1 is 3 , z 2 is z .

Substitute the value of x 1 , x 2 , y 1 , y 2 , z 1 and z 2 in equation (1),

AP= ( x−1 ) 2 + ( y−2 ) 2 + ( z−3 ) 2

To find the distance BP between the points B and P, the value of x 1 is 3 , x 2 is x , y 1 is 2 , y 2 is y , z 1 is −1 and z 2 is z .

Substitute the value of x 1 , x 2 , y 1 , y 2 , z 1 and z 2 in equation (1),

BP= ( x−3 ) 2 + ( y−2 ) 2 + ( z+1 ) 2

Since P is equidistant to both A and B, thus, AP=BP or AP 2 = BP 2 Therefore,

( x−1 ) 2 + ( y−2 ) 2 + ( z−3 ) 2 = ( x−3 ) 2 + ( y−2 ) 2 + ( z+1 ) 2 ( x 2 −2x+1 )+( y 2 −4y+4 )+( z 2 −6z+9 )=( x 2 −6x+9 )+( y 2 −4y+4 )+( z 2 +2z+1 ) x 2 − x 2 −2x+6x+ y 2 −4y− y 2 +4y= z 2 − z 2 +2z+6x−1−4−9+9+4+1 x=2z

Thus, the equation of the set of points which are equidistant from the points A=( 1,2,3 ) and B=( 3,2,−1 ) is given by x−2z=0