A bivariate Gaussian pdf is $$ f_{X, Y}(x, y)=\frac{1}{2 \pi|\Sigma|^{1 / 2}} \exp \left(-\frac{1}{2}\left[x-m_1, y-m_2\right] \Sigma^{-1}\left[x-m_1, y-m_2\right]^T\right) $$ where $\Sigma=\left[\begin{array}{cc}\sigma_1^2 & \rho \\ \rho & \sigma_2^2\end{array}\right]$ is the covariance matrix and vector $\left(m_1, m_2\right)$ is the mean vector.
How can I show that conditional pdf $f_{X \mid Y}(x \mid y)$ is a Gaussian pdf with mean $m_1+\frac{\rho}{\sigma_2^2}\left(y-m_2\right)$ and variance $\sigma_1^2-\frac{\rho^2}{\sigma_2^2}$?
I've tried using the fact that: $$f_{X|Y}(x|y) = \frac{f_{XY}(x,y)}{f_Y(y)} = \frac{f_{XY}(x,y)}{\int_{-\infty}^{\infty}f_{XY}(x,y) dx} $$
But I am getting lost in a mess of algebra. I want to know if there is a simpler way of going about this proof, and would really appreciate a detailed explanation.