Question 477701
First in an equilateral triangle all the sides are the same length

Let's call the triangle ABC  where A(4,0)  B(-6,0)  and C (x,y)

the sides of the triangle are AB, BC and AC
 
To find the length of a side of the Triangle we need to use the formula to calculate the distance between two points (x1, y1) and (x2, y2)

    Distance = {{{sqrt((x[1] -x[2])^2  + (y[1] - y[2])^2)}}}


Distance AB = {{{sqrt((4 - (-6))^2 + (0 - 0)^2)}}}

            = {{{sqrt((4 + 6)^2) + 0)}}}

            ={{{sqrt(10^2)}}} = 10

Since all sides are equal then length of BC = 10 and length of AC is also 10

Distance BC = {{{sqrt((-6-x)^2+(0 - y)^2)}}}

             = {{{sqrt((-6-x)^2 + y^2)}}}


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

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

BC = AC
{{{sqrt((-6-x)^2 + y^2)}}}  =   {{{sqrt((4-x)^2 + y^2)}}}

If we square both sides

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

Subtract {{{y^2}}} from both sides 

  {{{(-6-x)^2 }}}  =   {{{(4-x)^2 }}}

Expand

{{{36 + 12x + x^2   =  16 - 8x + x^2}}}

Subtract {{{x^2}}} from both sides 
{{{36 + 12x    =  16 - 8x }}}

Add 8x to both sides and subtract 36 from both sides

36 + 12x   =  16 - 8x
-36  +8x      -36  + 8x

20x  = - 20

Divide both sides by 20

 x = -1


So the x-coordinate is -1

Using the formula for length of BC  and plugging in the x-coordinate -1

Distance BC = {{{sqrt((-6-x)^2 + y^2)}}} = 10

              {{{sqrt((-6-(-1))^2 + y^2)}}}  = 10

              {{{sqrt((-6 + 1)^2 + y^2)}}}  = 10
              {{{sqrt((-5)^2 + y^2)}}}  = 10
              {{{sqrt(25 + y^2)}}}  = 10
Square both sides
              {{{(25 + y^2)}}}  = 100

 Subtract 25 from both sides
                          {{{y^2}}}  = 75
  Take square root of both sides

                y = +{{{sqrt(75)}}}  or y = -{{{sqrt(75)}}}

 y = +{{{5sqrt(3)}}}   or  y = -{{{5sqrt(3)}}}


The possible coordinates of the third vertex are (-1, +5{{{sqrt(3)}}}) and (-1, -5{{{sqrt(3)}}})