Question 682613
<pre>
Here is one side of the equilateral triangle:
 
{{{drawing(400,400,-6,4,-4,6, graph(400,400,-6,4,-4,6),locate(.1,.5,"(0,0)"),
 
line(0,0,-4,3) ,locate(-5.2,3.3,"(-4,3)") )}}}
 
And as you can see there are two solutions. Let the third
vertex be (a,b):
 
{{{drawing(400,400,-6,4,-4,6, graph(400,400,-6,4,-4,6),locate(.1,.5,"(0,0)"),
green(triangle(0,0,-4,3,.59807621,4.964102615)),locate(-5.2,3.3,"(-4,3)"),
locate(.58,5.2,"(a,b)"),
line(0,0,-4,3) ) }}}and{{{drawing(400,400,-6,4,-4,6, graph(400,400,-6,4,-4,6),
green(triangle(0,0,-4,3,-4.598078211,-1.964101615)),locate(.1,.5,"(0,0)"),
locate(-5.6,-1.9,"(a,b)"),
line(0,0,-4,3),locate(-5.2,3.3,"(-4,3)") ) }}}
 
First we find the distance from (0,0) to (-4,3) by the distance formula:
 
d = {{{sqrt( (x[2]-x[1])^2 + (y[2]-y[1])^2 ) }}}  
 
d = {{{sqrt((0-3)^2 + (0-(-4))^2 ) }}}
 
d = {{{sqrt((-3)^2 + (4))^2 ) }}}
d = {{{sqrt(9 + 16) }}}
d = {{{sqrt(25) }}}   
d = 5
 
So we know the distance from (0,0) to (a,b) must equal 5
so that the triangle will be equilateral.  So we use the
distance formula to make that equation:
 
{{{ sqrt((a-0)^2 + (b-0)^2 ) }}} = 5
{{{ sqrt(a^2 + b^2 ) }}} = 5
Square both sides:
a² + b² = 25
 
We also know the distance from (-4,3) to (a,b) must also
equal 5 so that the triangle will be equilateral.  So we 
use the distance formula again to make that equation:
 
{{{ sqrt((a-(-4))^2 + (b-3)^2 ) }}} = 5
{{{ sqrt((a+4)^2 + (b-3)^2 ) }}} = 5
Square both sides:
           (a+4)² + (b-3)² = 25
a² + 8a + 16 + b² - 6b + 9 = 25
    a² + 8a + b² - 6b + 25 = 25
         a² + 8a + b² - 6b = 0
 
So we have the system of equations:
 
          a² + b² = 25         
a² + 8a + b² - 6b = 0
 
Subtracting the second equation from the first equation
gives the equation:
 
-8a + 6b = 25
      6b = 25 + 8a
       b = {{{(25+8a)/6}}}
 
Substitute in
 
a² + b² = 25
 
a² + {{{( (25+8a)/6)^2}}} = 25
a² + {{{ (25+8a)^2/36}}} = 25

Multiply thru by 36 to clear of fractions:

36a² + (25 + 8a)² = 900

36a² + 625 + 400a + 64a² = 900

100a² + 400a - 275 = 0

Divide thru by 25

4a² + 16a - 11 = 0

Use the quadratic formula:

a = {{{(-16 +- sqrt( 16^2-4*4*(-11) ))/(2*4) }}}

a = {{{(-16 +- sqrt(256+176 ))/8 }}}

a = {{{(-16 +- sqrt(432 ))/8 }}}

a = {{{(-16 +- sqrt(144*3))/8 }}}

a = {{{(-16 +- 12sqrt(3))/8 }}}

a = {{{(4(-4 +- 3sqrt(3)))/8}}}

a = {{{(-4 +- 3sqrt(3))/2}}}

Substituting in

6b = 25 + 8a

6b = 25 + 8({{{(-4 +- 3sqrt(3))/2}}})

6b = 25 + 4(-4 +- 3{{{sqrt(3)}}})

6b = 25 - 16 +- 12{{{sqrt(3)}}})

6b = 9 ± 12{{{sqrt(3)}}})

Divide through by 3

2b = 3 ± 4{{{sqrt(3)}}}

b = {{{(3 +- 4sqrt(3))/2}}}

So the third coordinate has two solutions:

(a,b) = ({{{(-4 + 3sqrt(3))/2}}}, {{{(3 + 4sqrt(3))/2}}})

 and

(a,b) = ({{{(-4 - 3sqrt(3))/2}}}, {{{(3 - 4sqrt(3))/2}}})

These are approximately:

(a,b) = (0.598, 4.964)  and

(a,b) = (-4.598, -1.964)

Edwin</pre>