Question 442382


Find coordinates for the point equidistant from (2,1) (2,-4) (-3,1)


Let P(x, y) be the point.

The distance between (x, y) and (2, 1) is:

{{{D = sqrt((x - 2)^2 + (y -1)^2)}}}

The distance between (x, y) and (2, -4) is:

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


The distance between (x, y) and (-3, 1) is:

{{{D = sqrt((x +3)^2 + (y -1)^2)}}}

So we have the following system of equations in terms of x, y, and D:
{{{D = sqrt((x - 2)^2 + (y -1)^2)}}}
{{{D = sqrt((x - 2)^2 + (y +4)^2)}}}
{{{D = sqrt((x +3)^2 + (y -1)^2)}}}
 
Square through the square roots:
{{{D^2= (x - 2)^2 + (y -1)^2}}}
{{{D^2 = (x - 2)^2 + (y +4)^2}}}
{{{D^2 = (x +3)^2 + (y -1)^2}}}


Since these all equal D^2, we have two equations set equal to each other and we will produce a system of two equations that we are all used to.

Setting equation 1 = equation 2:

{{{ (x - 2)^2 + (y -1)^2=(x - 2)^2 + (y +4)^2}}}

{{{cross( (x - 2)^2 )+ (y -1)^2=cross((x - 2)^2)+ (y +4)^2}}}


{{{  (y -1)^2= (y +4)^2}}}

{{{  y ^2-2y+12= y^2+8y +16}}}

{{{  cross(y ^2)-2y+12= cross(y^2)+8y +16}}}

{{{  -2y+12= 8y +16}}}

{{{  -16+12= 8y +2y}}}

{{{  -4= 10y}}}


{{{  -4/10= y}}}

{{{  -2/5= y}}}...now find {{{x}}}


Setting equation 1 = equation 3:

{{{ (x - 2)^2 + (y -1)^2=(x +3)^2 + (y -1)^2}}}

{{{ (x - 2)^2 + cross((y -1)^2)=(x +3)^2 + cross((y -1)^2)}}}

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

{{{ x^2-4x +4 = x^2+6x +9 }}}

{{{ cross(x^2)-4x +4 =cross( x^2)+6x +9 }}}

{{{ -4x +4 = 6x +9 }}}


{{{ -9 +4 = 6x +4x }}}

{{{ -5= 10x }}}


{{{ -5/10= x }}}

{{{ -1/2= x }}}

so, the point is (-1/2, -2/5)