Question 989964
The two lines have y-intercepts equal to {{{-6}}}.
So the altitude of the equilateral triangle is {{{A=5-(-6)=11}}}
The altitude, half side, and side form a right triangle.
{{{A^2+(S/2)^2=S^2}}}
{{{A^2=(3/4)S^2}}}
{{{A=(sqrt(3)/2)S}}}
{{{11=(sqrt(3)/2)S}}}

{{{S=22/sqrt(3)}}}
The top edge is centered about {{{x=0}}} so the endpoints of the two vertices are 
({{{11/sqrt(3)}}},{{{5}}}) and ({{{-11/sqrt(3)}}},{{{5}}}).
Now that you have two points, find the slope.
{{{m=(5-(-6))/(11/sqrt(3)-0)}}}
{{{m=11/(11/sqrt(3))}}}
{{{m=sqrt(3)}}}
*[illustration fd8.JPG].