An equilateral triangle $ABC$ is inscribed in the parabola $y=x^2$ and one of the side of the equilateral triangle has the gradient 2. If the sum of $x-$-coordinates of the vertices of the triangle is a rational in the form $\frac{p}{q}$ where $p$ and $q$ are coprime, then find the value of $\frac{\left(\mathrm{p}+\mathrm{q}\right)}{7}$

y = x² 
To find t₁ + t₂ + t₃ = ?
${\mathrm{m}}_1=\frac{{\mathrm{t}}_2^2-{\mathrm{t}}_1^2}{{\mathrm{t}}_2-{\mathrm{t}}_1}={\mathrm{t}}_2+{\mathrm{t}}_1$
Similarly m₂ = t₂ + t₁ and m₃ = t₃ + t₁ 
Hence $\sum {\mathrm{t}}_{\mathrm{i}}=\frac{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3}{2}$
Now, $\tan {60}^{\circ }=\left|\frac{\mathrm{m}-2}{1+2\mathrm{m}}\right|$
$\pm \sqrt{3}(1+2\mathrm{m})=\mathrm{m}-2$
Taking + ve sign, we get $\sqrt{3}(1+2\mathrm{m})=\mathrm{m}-2$
$=\mathrm{m}(2\sqrt{3}-1)=-(2+\sqrt{3});\mathrm{m}=\frac{-(2+\sqrt{3})}{2\sqrt{3}-1}$
Taking – ve sign, we get
$$\begin{gathered} \begin{array}{l}\mathrm{m}-2=-2\sqrt{3}\mathrm{m}-\sqrt{3}\\ \mathrm{m}(2\sqrt{3}+1)=2-\sqrt{3};\mathrm{m}=\frac{2-\sqrt{3}}{2\sqrt{3}+1}\\ {\mathrm{m}}_1=\frac{-(2+\sqrt{3})}{2\sqrt{3}-1};{\mathrm{m}}_2=\frac{2-\sqrt{3}}{2\sqrt{3}+1}\text{ and }{\mathrm{m}}_3=2\\ =\sum _{\mathrm{i}=1}^3 {\mathrm{t}}_{\mathrm{i}}=\frac{{\mathrm{m}}_1+{\mathrm{m}}_2+{\mathrm{m}}_3}{2}=\frac{\frac{-(2+\sqrt{3})}{2\sqrt{3}-}+\frac{2-\sqrt{3}}{2\sqrt{3}+1}}{2}\\ =\frac{\frac{-4\sqrt{3}-2-6-\sqrt{3}+4\sqrt{3}-6-2+\sqrt{3}+22}{11}}{2} \\ =\frac{6}{22}=\frac{3}{11}=\frac{\mathrm{p}}{\mathrm{q}}=(\mathrm{p}+\mathrm{q})=14\end{array} \end{gathered}$$