Conditioning of join gaussian over a line I need to compute the conditional probability of bivariate normal distribution over a line.
Let's suppose that X and Y both are normal distributions and that they are independent.
Let's suppose that we want to find the probability that a point 'k' will be on the line  L= aX+bY=c.
To do so we need to compute the following: 
$ p(k|aX+bY=c) = \frac {p_x(X_k)p_y(Y_k)}{p(aX+bY=c)} = \frac {p_x(X_k)p_y(Y_k)}{\int_{L} p_x(X)p_y(Y) dX dY}$
where k is point over the line L.
For this to happen we will use a base for the line L define as a vector b(b1,b2) and a point of origin r(r1,r2). Therefore, the previous integral can be written as: 
$ \frac {p_x(X_k)p_y(Y_k)}{\int_{t} p_x(r_1+b_1t)p_y(r_2+b_2t) dt}  = \frac {p_x(r_1+b_1k)p_y(r_2+b_2k)}{\int_{t} p_x(r_1+b_1t)p_y(r_2+b_2t) dt}$
My goal now is to show that this last fraction is indeed a gaussian distribution and to get its mean and variance

A bivariate normal density can be likened to a piece of bologna (or did I mean to write baloney?) about which Americans often say "No matter how you slice it, it is still bologna". The meaning w.r.t. bivariate normal densities (regarded as a piece of bologna sitting on the $x$-$y$ plane) is that every cross-section of this piece (of lunch meat) has the shape of a univariate normal density. Slice it along the straight line $ax+by=c$ and the cross-section has the shape of a normal density.  It is not exactly a univariate normal density in that the "area under the curve" in usually not $1$ as it must be for all valid univariate densities, but for any nonnegative function $g(x)$ with finite area $A$, $A^{-1}g(x)$ is a valid density function.  So, as I said, the shape is of the correct form, but the area need not be, but the area is readily fixed.
So, what is the density of $aX+bY+c$ when $(X,Y)$ is bivariate normal? Easy-peasy -- it is a normal density with mean $$E[aX+bY+c] = a\mu_x +b\mu_Y +c$$ and variance
$$\operatorname{var}(aX+bY+c) = a^2\operatorname{var}(X) + b^2\operatorname{var}(Y) +2ab\operatorname{cov}(X,Y).$$
