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2014-08-19T23:32:33+05:30
Let say the point P(x, y) is on the locus. Now (x, y) is at the same distance from (3, 7) and (-2, 4).

Formula :- Distance between two points (a, b) and (c, d) = \sqrt{ (a - b)^{2} +  (c - d)^{2} }

Distance between (x, y) and (3, 7) = \sqrt{ (x - 3)^{2} +  (y - 7)^{2} }

Distance between (x, y) and (-2, 4) = \sqrt{ (x + 2)^{2} +  (y - 4)^{2} }

Now,
\sqrt{ (x - 3)^{2} +  (y - 7)^{2} } = \sqrt{ (x + 2)^{2} +  (y - 4)^{2} } \\  \\ (\sqrt{ (x - 3)^{2} +  (y - 7)^{2} })^{2} = (\sqrt{ (x + 2)^{2} +  (y - 4)^{2} })^{2} \\  \\  x^{2} - 6x + 9 +  y^{2} - 14y + 49 =  x^{2} + 4x + 4 +  y^{2} - 8y + 16 \\  \\ -6x -14y + 58 = 4x - 8y + 20 \\ 10x + 6y = 38 \\ 5x + 3y = 19

Answer
5x + 3y = 19

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2014-08-20T00:00:24+05:30

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Find middle point P of the two  given points : [ (3-2)/2 , (7+4)/2  ]
             = ( 0.5 , 5.5 )
The locus passes through this.  The locus is a straight line bisecting perpendicularly the line joining the given points.  So find slope of the line joining given points.  
        slope =  (7- 4) / (3 +2 )  = 3/5 = 0.6
Now the perpendicular bisector has as slope of  - 1/0.6
So equation of the locus desired :
         (y - y1)    =  m (x - x1)
         y - 5.5 = - (x -0.5) / 0.6
         0.6 y - 5.5*6 = - x  + 0.5 

           0.6 y + x = 3.8      or  3 y + 5 x = 19


 
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