= Solution
For common <positive-definite> covariance, subtracting the two Gaussian log densities cancels the quadratic term. Equal priors leave the <linear discriminant analysis> score
$$
\delta(x)=(\mu_1-\mu_2)^T\Sigma^{-1}\left(x-\frac{\mu_1+\mu_2}{2}\right).
$$
Here
$$
\Sigma^{-1}=\begin{pmatrix}1/2&-1\\-1&3\end{pmatrix},\qquad
\Delta=\mu_1-\mu_2=\begin{pmatrix}3\\-1\end{pmatrix},\qquad
\Sigma^{-1}\Delta=\begin{pmatrix}5/2\\-6\end{pmatrix}.
$$
Thus
$$
\boxed{\delta(x)=\tfrac52x_1-6x_2+\tfrac{31}4;\quad
\text{assign class one if }10x_1-24x_2+31>0.}
$$
Assign class two if the expression is negative; a tie has zero probability under either continuous Gaussian law.
To compute the error rather than just specify the boundary, put $D^2=\Delta^T\Sigma^{-1}\Delta=27/2$. Within class one the score has mean $D^2/2=27/4$, and within class two mean $-27/4$. In both classes its <variance> is $\Delta^T\Sigma^{-1}\Sigma\Sigma^{-1}\Delta=D^2$. Hence both conditional error probabilities are $\Phi(-D/2)$, where $\Phi$ is the <standard normal distribution> function. Equal priors give the <equal-covariance Gaussian classification error>
$$
\boxed{P_{\mathrm{error}}=\Phi\left(-\sqrt{27/8}\right)\approx0.0331.}
$$
For the specified observation, the score is $\delta(2,1)=27/4>0$, so \b[assign it to class one].
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