Question 1160443
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<pre>

Let (x,y) be the current point on the perpendicular bisector.


Its distance from the point (4,3) is  {{{sqrt((x-4)^2+(y-3)^2)}}}.


Its distance from the point (-2,5) is  {{{sqrt((x+2)^2+(y-5)^2)}}}.


The distances are equal

    {{{sqrt((x-4)^2+(y-3)^2)}}} = {{{sqrt((x+2)^2+(y-5)^2)}}}.


Square both sides

    (x-4)^2 + (y-3)^2 = (x+2)^2 + (y-5)^2.


Simplify

    x^2 - 8x + 16 + y^2 - 6y + 9 = x^2 + 4x + 4 + y^2 - 10y + 25

    -8x - 4x + (-6y + 10y) = 4 + 25 - 16 - 9

    -12x + 4y = 4

     3x - y = -1.     <U>ANSWER</U>
</pre>

Solved.