Integral of the product of two gaussian Here is the big picture of my problem:
In the image below, X and Y represent 2 independent gaussian distributions. So the circles are the representation of the resultant bivariate gaussian. This bivariate gaussian is traversed by a line L: aX+bY=c where a,b and c are real numbers. So It is know that when a bivariate gaussian is traversed by a line (so in 3 d this bivariate gaussian is sliced by a plane) the result is a gaussian distribution.
https://i.stack.imgur.com/vY6jr.png
In order to derive that analytically, we compute the conditional distribution of the bivariate gaussian over the line L.
This is done by:
$ p(k|aX+bY=c) = \frac {p_x(X_k)p_y(Y_k)}{p(X+Y=c)} = \frac {p_x(X_k)p_y(Y_k)}{\int_{L} p_x(X)p_y(Y) dX dY}$
Where $k$ defines any point on the line L
The numerator is easy since it is the product of gaussian which will give an unnormalized gaussian.
However, my problem is the denominator. Indeed, I know for sure that this integral in the denominator is equal to $P_z(c)$ where P_z is the gaussian distribution : $Z=aX+bY$.
My goal is to prove that result (the one related to the integral in the denominator)
I guess that to approach that integral, first we need to do a change of base using the base from the line which is characterized by a point of origin $(r_1,r_2)$ and $(b_1,b_2)$

This is the third time you've asked a similar question, and I think the problem is you're making a fundamental misunderstanding between discrete probability and continuous probability. That is leading you to a denominator with an integral that equals 0.
For discrete probability I think you understand, but I'll describe it to be clear: Let $X$ and $Y$ be two random variables, which may or may not be independent. The conditional probability of $Y$ given $X = x$ is
$$P(Y=y|X=x) = \frac{P(Y=y , X=x)}{P(X=x)}$$
(assuming $P(X=x) > 0$). For independent $X$ and $Y$, we have $P(Y=y , X=x) = P(Y=y)P(X=x)$, so the final result is $P(Y=y|X=x) = P(Y=y)$, as we should expect.
For continuous probability it is similar, except we talk about the density of the random variables, not the probability. Let $f_{X,Y}(x,y)$ be the joint density of $X$ and $Y$ at the point $(x,y)$, $f_X(x)$ be the density of $X$, $f_Y(y)$ be the density of $Y$, and $f_{Y|X=x}(y)$ be the conditional density of $Y$ given $X=x$. Also assume that $f_X(x) > 0$. The expression for the conditional density of $f_{Y|X=x}(y)$ is
$$
f_{Y|X=x}(y) = \frac{f_{X,Y}(x,y)}{f_X(x)}
$$
In general $f_{Y|X=x}(y)$ will be a function of both $x$ and $y$. Once again you will notice that if $X$ and $Y$ are independent, then $f_{X,Y}(x,y) = f_X(x)f_Y(y)$, and we end up with $f_{Y|X=x}(y) = f_Y(y)$, as we should expect. Also note that there is no integration in the denominator of this expression.
So, to find the conditional density of $K | aX + bY = c$, you need to find the joint density of $K$ and $aX + bY$ and the density of $aX + bY$. This answer to your previous question used properties of the joint normal distribution to find the joint distribution of $X$ and $aX + bY$, then found the conditional distribution of $X|aX+bY=c$, and then used the relation in the definition of the line to transform $X$ into $K$. This avoided working with the densities directly.
Proving the result directly from the densities will involve some unpleasant algebra, but should follow naturally as described in this question, Deriving the conditional distributions of a multivariate normal distribution, now that you know to get rid of the integral sign in the denominator of your expression.
