Question

find the locus of a point which is equidistant from the point (1,2,3) and (3,2,1)

Answer 1

Given,
Lines are
 x+y+4 = 0 and 7x + y + 20 = 0
In line,
x+y+ 4 = 0
By putting x = -3 and y = -1
LHS 
=(-3)+(-1)+4 
=-4+4 
=0
RHS
=0

LHS = RHS

Hence, point (-3,-1) lies in the line
In line,
7x + y + 20 = 0
By putting x = -2 and y = -6
LHS   
=7(-2)+(-6)+20 
=-14-6+20
=0

RHS
=0

LHS = RHS
Hence, point (-2,-6) lies in the line
             m1 = 1        m2=1 
  --------------------.------------------------
 A(-3,-1)            P(x,y)                    B(-2,-6) 

By section formula,
x,y = (m1x2 + m2x1/ m1+m2 , m1y2 + m2y1/ m1+m2)
      = [1*(-2) + 1 * (-3) / 1+1     ,  1 * -6  + 1 * -1  /1+1]
      =[-2-3 / 2 , -6 - 1 /2)
      =(-5/2 , -7/2)
Hence P(-5/2 , -7/2) is the point of equidistance from points (-3,-1)  of the line x+y+4 = 0 and(-2,-6) of the line 7x+y+20=0.