Weighted averaging of polynomial coefficients Suppose there are three polynomials $$y_1 = a_0 + a_1x_1, $$ $$y_2 = b_0 + b_1x_2, $$ $$y_3 = c_0 + c_1x_3,$$

and define $$ x_{\rm avg} \triangleq \dfrac{x_1(x_2-x_3)^2+x_2(x_1-x_3)^2+x_3(x_1-x_2)^2}{(x_1-x_2)^2+(x_1-x_3)^2+(x_2-x_3)^2}, $$
$$ y_{\rm avg} \triangleq \dfrac{y_1(y_2-y_3)^2+y_2(y_1-y_3)^2+y_3(y_1-y_2)^2}{(y_1-y_2)^2+(y_1-y_3)^2+(y_2-y_3)^2}. $$

I want to find $d_0 = f(a_0,b_0,c_0,a_1,b_1,c_1)$ and $d_1 = f(a_0,b_0,c_0,a_1,b_1,c_1)$ such that $$y_{\rm avg} = d_0 + d_1x_{\rm avg}.$$

Is this possible? I would guess yes, but I'm not really sure. If it is, how should I solve for $d_0,d_1$ and does the solution extend to higher order polynomials? I've tried substituting the polynomials $y_1,y_2,y_3$ into the $y_{\rm avg}$ equation, but I'm unsure how to extract meaningful coefficients that are independent of $x_1,x_2,x_3$.
Here is a counterexample. Assuming $x_{\rm avg} = M$ on the diagonal $x_1 = x_2 = x_3 = M$, then $y_{\rm avg} = d_0 + d_1 x_{\rm avg} = d_0 + d_1 M$, for some fixed $M$. For simplicity, fix $a_0 = 1, b_0 = 0, c_0 = 0$ and $a_1 = 1, b_1 = 2, c_1 = 3$. 

If $M = 0$ then $y_{\rm avg} = 0 = d_0$. If $M = 1$ then $y_{\rm avg} = 2 = d_1$, but if $M = 2$ then $y_{\rm avg} \approx 3.86 \neq 2d_1$.
Here is a counterexample. Assuming $x_{\rm avg} = M$ on the diagonal $x_1 = x_2 = x_3 = M$, then $y_{\rm avg} = d_0 + d_1 x_{\rm avg} = d_0 + d_1 M$, for some fixed $M$. For simplicity, fix $a_0 = 1, b_0 = 0, c_0 = 0$ and $a_1 = 1, b_1 = 2, c_1 = 3$. 

If $M = 0$ then $y_{\rm avg} = 0 = d_0$. If $M = 1$ then $y_{\rm avg} = 2 = d_1$, but if $M = 2$ then $y_{\rm avg} \approx 3.86 \neq 2d_1$.