I'm reading Merton's An Analytic Derivation of the Efficient Portfolio Frontier. In section IV, he derives the efficient frontier with a riskless asset. Let $\mathbf{w}$ be a vector of portfolio weights and let $w_f$ be the weight of the risk-free asset $r_f$. Then
$$ \mathbf{w}^{\top} \mathbf{1} + w_f = 1 \tag{1} $$
by construction. The optimization problem is
$$ \begin{aligned} \min_{\mathbf{w}} &&& \mathbf{w}^{\top} \boldsymbol{\Sigma} \mathbf{w}, \\ \text{subject to} &&& \mathbf{w}^{\top} \tilde{\boldsymbol{\mu}} = \tilde{\mu}_p, \end{aligned} \tag{2} $$
where
$$ \begin{aligned} \tilde{\boldsymbol{\mu}} &\triangleq \boldsymbol{\mu} - r_f \mathbf{1}, \\ \tilde{\mu}_p &\triangleq \mu_p - r_f, \end{aligned} \tag{3} $$
and where $\boldsymbol{\mu}$ is a vector of expected returns and $\mu_p$ is the portfolio's return. I can write down the Lagrangian function and derive the first-order conditions:
$$ \begin{aligned} \nabla_{\mathbf{w}} \mathcal{L} &= 2 \boldsymbol{\Sigma} \mathbf{w} + \lambda \tilde{\boldsymbol{\mu}} = \mathbf{0}, \\ \frac{\partial}{\partial \lambda} \mathcal{L} &= \mathbf{w}^{\top} \tilde{\boldsymbol{\mu}} - \tilde{\mu}_p = 0. \end{aligned} \tag{4} $$
Finally, I can derive the same optimal weights
$$ \mathbf{w} = \tilde{\mu}_p \left( \frac{\boldsymbol{\Sigma}^{-1} \tilde{\boldsymbol{\mu}}}{\tilde{\boldsymbol{\mu}}^{\top} \boldsymbol{\Sigma}^{-1} \tilde{\boldsymbol{\mu}}} \right) \tag{5} $$
and the same quadratic-form equation as Merton, his equation 35:
$$ \begin{aligned} | \mu_p - r_f | &= \sigma_p \sqrt{(\boldsymbol{\mu} - r_f \mathbf{1})^{\top} \boldsymbol{\Sigma}^{-1} (\boldsymbol{\mu} - r_f \mathbf{1})} \\ &\Downarrow \\ \mu_p &= r_f \pm \sigma_p \sqrt{(\boldsymbol{\mu} - r_f \mathbf{1})^{\top} \boldsymbol{\Sigma}^{-1} (\boldsymbol{\mu} - r_f \mathbf{1})}. \end{aligned} \tag{6} $$
This is clearly a piecewise function in which each half is a linear function. The top half is, I assume, what people call the capital market line, since the independent variable is $\mu_p$, the dependent variable is $\sigma_p$, and the $y$-intercept is $r_f$. However, and this is my question, the slope is not the Sharpe ratio:
$$ \frac{\mu_p - r_f}{\sigma_p} \stackrel{???}{\neq} \sqrt{(\boldsymbol{\mu} - r_f \mathbf{1})^{\top} \boldsymbol{\Sigma}^{-1} (\boldsymbol{\mu} - r_f \mathbf{1})}. \tag{7} $$
What am I missing?