Question 1192225
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Find the equation of the locus of all points (x,y) that are equidistant from (-3, 4) and (0, -3).


<B>Solution</B> 


<pre>
Let (x,y) be the point of the locus.


Then the equation expressing equidistance from the points (-3, 4) and (0, -3) is


    {{{sqrt((x-(-3))^2+(y-4)^2)}}} = {{{sqrt((x-0)^2+(y-(-3))^2)}}}


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


    {{{x^2+6x+9 + y^2-8y+16}}} = {{{x^2 + y^2+6y+9}}}


    6x - 8y + 25 = 6y + 9


    6x - 14y = 9 - 25


    6x - 14y = -16


    3x -  7y =  -8


<U>ANSWER</U>.  We get the equation of the straight line  3x - 7y = -8  in the general form.
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Solved.