= Solution
Use base-ten <logarithms> and dimensionless <thermodynamic activities> at the stated temperature and pressure. Define $x=\log_{10}a_{\mathrm{H_4SiO_4}}$ and $y=\log_{10}(a_{\mathrm{Na^+}}/a_{\mathrm{H^+}})$. The concentration notation is an ideal dilute-solution approximation with a consistent concentration reference.
With pure solids and unit water activity, the two <reaction quotients> are $10^{2y+2x}$ and $10^{2y+6x}$. Equating them to the corresponding <equilibrium constants> gives
$$
\boxed{y=3.44-x,\qquad y=-7.59-3x.}
$$
The slopes on the <mineral activity diagram> are $-1$ and $-3$. The lines meet at $(x_*,y_*)=(-5.515,8.955)$, within the negative silica-logarithm range.
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