Let's say I hypothesize that a graph of $\ln K$ vs. $1/T$ has a slope of $-∆G^\circ/R$ and a $y$-intercept of $0$. I prove it simply:
$$∆G^\circ = -RT\ln K \quad\to\quad \ln K = -\frac{∆G^\circ}{RT}$$
This matches the linear form $y = mx + b$. Thus, plotting $\ln K$ vs. $1/T$ would have a slope $m = -∆G^\circ/R$ and a $y$-intercept $b = 0$.
However, I understand that a van't Hoff plot defines a graph of $\ln K$ vs. $1/T$ to have a slope of $-ΔH^\circ/R$ and a $y$-intercept of $∆S^\circ/R$. It is clear from the relation $∆G^\circ = ∆H^\circ - TΔS^\circ$ that my final equation is thermodynamically equivalent to the van't Hoff equation. I do not disagree that
$$\ln K = -\frac{∆H^\circ}{RT} + \frac{∆S^\circ}{R},$$
but if I were to experimentally measure temperature and calculate the equilibrium constant temperature, why should I expect the y-intercept to be $∆S^\circ/R$ as defined by van't Hoff rather than $0$ as I defined above? Why should I expect the slope to be $-ΔH^\circ/R$ instead of $-ΔG^\circ/R$? What makes the van't Hoff equation match experimentally determined values over the equation $\ln K = -∆G^\circ/(RT)$?