Conditional expectation of a joint normal distribution

I've found an answer that I'm happy with:

$$ Y - \frac{\mathrm{cov}(X, Y)}{\mathrm{var}(X)} X $$ is jointly normal with $X$ and uncorrelated, hence independent.

Therefore

$$\mathbb{E}[Y|X] = \mathbb{E}[Y - \frac{\mathrm{cov}(X, Y)}{\mathrm{var}(X)} X + \frac{\mathrm{cov}(X, Y)}{\mathrm{var}(X)} X| X] \\ = \mathbb{E}[Y - \frac{\mathrm{cov}(X, Y)}{\mathrm{var}(X)} X] +\frac{\mathrm{cov}(X, Y)}{\mathrm{var}(X)}X \\ = \mathbb{E} [Y] + \frac{\mathrm{cov}(X, Y)}{\mathrm{var}(X)} (X - \mathbb{E}[X])$$

Let $(X_1, X_2) \sim MVN (\mu, \Sigma)$, then recall that \begin{align} f_{X_2|X_1}(x_2) = \frac{f_{X_1, X_2}(x_1, x_2)}{f_{X_1}(x_1)} \, , \end{align} where $$ f_{X_1,X_2}(x_1,x_2)=\\ \frac{1}{2\pi \sigma_1\sigma_2 \sqrt{1-\rho}}\exp\left( - \frac{1}{2(1-\rho^2)}\left(\frac{(x_1 - \mu_1)^2}{\sigma^2} + \frac{(x_2 - \mu_2)^2}{\sigma^2} - \frac{2\rho(x_1-\mu_1)(x_2 - \mu_2)}{\sigma_1\sigma_2} \right) \right), $$ and $$ f_{X_1}(x) = \frac{1}{\sqrt{2\pi\sigma_1^2}}\exp\left( - \frac{(x_1-\mu_1)^2}{2\sigma_1^2} \right). $$

After some simple algebra and rearrengments you'll find that $$ X_2|X_1 = x_1 \sim N\left( \mu_2 + \rho\frac{\sigma_2}{\sigma_1}(x_1 - \mu_1), (1- \rho^2)\sigma_2^2 \right). $$