Question 938101
Two vertices of an equilateral triangle are (3,4)and (-2,3) the third vertex can be
<pre>
There are obviously two solutions:

{{{drawing(1800/11,200,-4,5,-2,9, graph(1800/11,200,-4,5,-2,9),
locate(3,4,"(3,4)"), locate(-4,3,"(-2,3)"),
line(3,4,-2,3), line((1-sqrt(3))/2,(7+5sqrt(3))/2,-2,3),
line((1-sqrt(3))/2,(7+5sqrt(3))/2,3,4),locate(-2.7,8.1,"(u,v)"),
line(-4,-2,5.3,-2),line(-4,-2,-4,9),line(-4,9,5.3,9),line(5.3,9,5.3,-2)
  )}}} and {{{drawing(1800/11,200,-4,5,-2,9, graph(1800/11,200,-4,5,-2,9),
locate(3,4,"(3,4)"), locate(-4,3,"(-2,3)"),
line(3,4,-2,3), line((1+sqrt(3))/2,(7-5sqrt(3))/2,-2,3),
line((1+sqrt(3))/2,(7-5sqrt(3))/2,3,4), locate(1,-1,"(u,v)"),
line(-4,-2,5.3,-2),line(-4,-2,-4,9),line(-4,9,5.3,9),line(5.3,9,5.3,-2)

  )}}}

First we find the distance between them to find out the
lengths of the three sides of the equilateral triangle.
{{{d = sqrt((x[2]-x[1])^2+(y[2]-y[1])^2)}}}
{{{d = sqrt((-2-3)^2+(3-4)^2)}}}
{{{d = sqrt((-5)^2+(-1)^2)}}}
{{{d = sqrt(25+1)}}}
{{{d = sqrt(26)}}}

Let the third vertex be the point (u,v)
 
Then the distance from (u,v) to (3,4) must
equal to the distance from (u,v) to (-2,3),
and each must equal to {{{sqrt(26)}}}.
{{{sqrt((u-3)^2+(v-4)^2)}}}{{{""=""}}}{{{sqrt((u-(-2)^"")^2+(v-3)^2)}}}{{{""=""}}}{{{sqrt(26)}}}
Squaring all three sides:
{{{(u-3)^2+(v-4)^2}}}{{{""=""}}}{{{(u+2)^2+(v-3)^2}}}{{{""=""}}}{{{26}}}
{{{(u-3)^2-(u+2)^2}}}{{{""=""}}}{{{(v-3)^2-(v-4)^2}}}
{{{((u-3)^""-(u+2))((u-3)^""+(u+2))}}}{{{""=""}}}{{{((v-3)^""-(v-4))((v-3)+(v-4))}}}   
{{{(u-3-u-2)(u-3+u+2)}}}{{{""=""}}}{{{(v-3-v+4)(v-3+v-4)}}} 
{{{(-5)(2u-1)}}}{{{""=""}}}{{{(1)(2v-7)}}}
{{{-10u+5}}}{{{""=""}}}{{{2v-7}}}
{{{-10u-2v}}}{{{""=""}}}{{{-12}}}
{{{5u+v}}}{{{""=""}}}{{{6}}}
{{{v=6-5u}}}
Substitute in
{{{(u+2)^2+(v-3)^2}}}{{{""=""}}}{{{26}}}
{{{(u+2)^2+(6-5u-3)^2}}}{{{""=""}}}{{{26}}}
{{{(u+2)^2+(3-5u)^2}}}{{{""=""}}}{{{26}}}
{{{u^2+4u+4+9-30u+25u^2=26}}}
{{{26u^2-26u+13=26}}}
{{{26u^2-26u-13=0}}}
{{{2u^2-2u-1=0}}}  
{{{u = (-b +- sqrt( b^2-4ac ))/(2a) }}} 
{{{u = (-(-2) +- sqrt( (-2)^2-4(2)(-1) ))/(2*2) }}}
{{{u = (2 +- sqrt(4+8))/4 }}}
{{{u = (2 +- sqrt(12))/4 }}}
{{{u = (2 +- sqrt(4*3))/4 }}}
{{{u = (2 +- 2sqrt(3))/4 }}}
{{{u = (2(1 +- sqrt(3)))/4 }}}
{{{u = (1 +- sqrt(3))/2 }}}
----
Using the +,
{{{u = (1 + sqrt(3))/2 }}}
{{{v=6-5u}}}
{{{v=6-5((1 + sqrt(3))/2)}}}
{{{v=12/2-(5 + 5sqrt(3))/2)}}}
{{{v=(12-(5 + 5sqrt(3)))/2)}}}
{{{v=(12-5 - 5sqrt(3))/2}}}
{{{v=(7 - 5sqrt(3))/2}}},

so one solution is {{{(matrix(1,3,(1 + sqrt(3))/2,",",(7 - 5sqrt(3))/2))}}}, which is about (-0.4,7.8) 

----

Using the -,
{{{u = (1 - sqrt(3))/2 }}}
{{{v=6-5u}}}
{{{v=6-5((1 - sqrt(3))/2)}}}
{{{v=12/2-(5 - 5sqrt(3))/2)}}}
{{{v=(12-(5 - 5sqrt(3)))/2)}}}
{{{v=(12-5 + 5sqrt(3))/2}}}
{{{v=(7 + 5sqrt(3))/2}}},

so the other solution is {{{(matrix(1,3,(1 + sqrt(3))/2,",",(7 + 5sqrt(3))/2))}}}, which is about (1.4,-0.8) 
 
Edwin</pre>