Question 938100
We know one side of the triangle, connecting vertices {{{A(3,4)}}} and {{{B(-2,3)}}} .
The length of  side AB is
{{{sqrt((3-(-2))^2+(4-3)^2)=sqrt((5)^2+(1)^2)=sqrt(25+1)=sqrt(26)}}}
Side AB is a segment of a line with slope
{{{m=(4-3)/(3-(-2))=1/5}}}.
The midpoint of AB, point P, has coordinates
{{{x[P]=(3+(-2))/2=1/2}}} and {{{y[P]=(4+3)/2=7/2}}} .
The equilateral triangles with side AB are ABD and ABC,
with C and D on opposite sides of segment AB.
Visualizing the location of vertices C and D is easy,
but I could not think of a simple way to calculate their coordinates.
 
{{{drawing(210,300,-3,4,-1,9,
grid(0),locate(0.6,3.5,P),
green(triangle(3,4,1.366,-0.83,-0.366,7.83)),
green(triangle(-2,3,-0.366,7.83,1.366,-0.83)),
green(line(0.4,4,0.9,4.1)),green(line(1,3.6,0.9,4.1)),
line(3,4,-2,3),locate(3,4,A),locate(-2.3,3,B),
locate(1.5,-0.5,C),locate(-0.7,8,D)
)}}}
The perpendicular bisector of AB contains the altitude of those triangles.
That perpendicular bisector is a line perpendicular to AB,
and passing through point P (the midpoint of AB).
Being perpendicular to a line with slope {{{m=1/5}}} ,
that altitude is part of a line with slope {{{-1/m=-5}}} .
 
{{{PA=(1/2)*AB=sqrt(26)/2}}} .
Since angle PAC is a {{{60^o}}},
{{{tan(PAC)=PC/PA=sqrt(3)}}}-->{{{PC=sqrt(3)*PA=sqrt(3)*sqrt(26)/2=sqrt(78)/2}}}
So the length of PC and PD is {{{PC=PC=sqrt(78)/2}}} ,
but what are the coordinates of C and D?
 
{{{drawing(210,390,-3,4,-2,11,
locate(0.6,3.5,P),green(arrow(0.5,-2,0.5,15)),
locate(0.6,10.8,green(x=1/2)),
green(triangle(0.5,3.5,1.366,-0.83,0.5,-0.83)),
green(triangle(0.5,3.5,0.5,7.83,-0.366,7.83)),
green(rectangle(0.5,-0.83,0.7,-0.63)),
green(rectangle(0.5,7.83,0.3,7.63)),
line(3,4,-2,3),locate(3,4,A),locate(-2.3,3,B),
locate(1.5,-0.5,C),locate(-0.7,8,D),
locate(-0.1,8.3,green(a)),locate(0.8,-0.7,green(a)),
locate(0.6,5.9,green(b)),locate(0.6,1.4,green(b))
)}}} Since the slope of CD is {{{-5}}} , {{{b=5a}}} ,
and in those two green right triangles
{{{a^2+b^2=a^2+(5a)^2=a^2+25a^2=26a^2=(sqrt(78)/2)^2=78/4}}}--->{{{a^2=78/4*26=3/4}}}--->{{{a=sqrt(3)/2}}}--->{{{b=5sqrt(3)/2}}}
Then,
{{{x[C]=1/2+sqrt(3)/2=(1+sqrt(3))/2=about1.37}}} (rounded),
{{{y[C]=7/2-5sqrt(3)/2=(7-5sqrt(3))/2=about-0.83}}} (rounded),
{{{x[D]=1/2-sqrt(3)/2=(1-sqrt(3))/2=about-0.37}}} (rounded), and
{{{y[D]=7/2+5sqrt(3)/2=(7+5sqrt(3))/2=about7.83}}} (rounded).