Uniform distribution on the triangle $ ∆:=\left \{ (x, y) ∈ R^2|0 < x, y < 1, x + y < 1 \right \}$ I have one small question?
Let $(X, Y )$ be a random vector on a probability space $(Ω, A, P)$
with the uniform distribution on the triangle $ ∆:=\left \{  (x, y) ∈ R^2|0 < x, y < 1, x + y < 1 \right \}$
I have to find the distribution function and probability density of:
a)$\frac{X}{Y}$
b)$X+Y$
for a) we had defined
$f_{X|Y}(X|Y):= \left\{\begin{matrix}
\frac{f(x,y)}{f_y(y)} & f_y(y)>0 \\ 
 0 & otherwise 
\end{matrix}\right.$
having already calculated $f(x,y)$ and $f_y(y)$ my result is this:
$f_{X|Y}(X|Y):= \left\{\begin{matrix}
\frac{1}{1-y} & f_y(y)>0 \\ 
 0 & otherwise 
\end{matrix}\right.$
It should be the right result.
Knowing that $f_{X|Y}(X|Y)=F'_{X|Y}(X,Y)$ I calculated $F(X,Y)=\int f_{X|Y}dx $
obtaining therefore the result:
$F_{X|Y}(X|Y):= \left\{\begin{matrix}
\ -ln(|y-1|)& f_y(y)>0 \\ 
 0 & otherwise 
\end{matrix}\right.$
However the corrections say that my result is wrong and I just can't figure out where I went wrong.
Can anyone help me?
 A: Let $Q= X/Y$ and $S=X+Y$ ($Q$ for quotient and $S$ for sum). Personally, for continuous random variables, I find it easier to find the distribution before the density function.
For $Q$: Since $X,Y\geqslant 0$, if $q\leqslant 0$ then $\operatorname{Pr}(Q\leqslant q)=0$, so we may assume that $q>0$. Then \begin{align*}
\operatorname{Pr}(Q\leqslant q) &= \operatorname{Pr}(X\leqslant qY) \\
&= \iint_D f_{(X,Y)}(x,y) \, dy \, dx \\
\end{align*} where $D = \{ (x,y ) \in [0,1]^2 \text{ s.t. } x+y<1, \, x\leqslant q y\}$. I've included a diagram of this region below.

Since $f_{(X,Y)}(x,y)=2$ in $\Delta$, \begin{align*}
\operatorname{Pr}(Q\leqslant q) &= 2\iint_D  \, dy \, dx \\
&= 2\int_0^{\frac q {q+1}} \big ( 1-x-\frac 1 q x \big ) \, dx \\
&= 2\bigg ( \frac q {q+1} - \frac 12\bigg (1 + \frac 1 q \bigg ) \bigg ( \frac q {q+1} \bigg )^2\bigg )\\
&= \frac q{q+1}.
\end{align*} Thus, $$f_Q(q) = \frac{d}{dq} \operatorname{Pr}(Q\leqslant q) = \begin{cases}
0, &\text{if }q< 0 \\
\frac 1 {(q+1)^2}, &\text{if } q>0.
\end{cases}$$
For $S$: Again, since $X,Y \geqslant 0$, if $s\leqslant 0$ then $\operatorname{Pr}(S\leqslant s)=0$. Moreover, $X+Y<1$, so if $s\geqslant 1$ then $\operatorname{Pr}(S\leqslant s)=1$. Assume that $0<s<1$. Then \begin{align*}
\operatorname{Pr}(S\leqslant s) &= 2 \int_0^s (s-x) \, dx \\
&=  s^2.
\end{align*} Again, I've included a diagram below.

Thus, $$\operatorname{Pr}(S\leqslant s) = \begin{cases}
0, &\text{if } s<0 \\
 s^2, &\text{if } 0<s<1 \\
1, &\text{if } s\geqslant 1 
\end{cases} $$ and $$F_S(s)= \frac{d}{ds} \operatorname{Pr}(S\leqslant s) =\begin{cases}
0, &\text{if } s<0 \text{ or } s>1 \\
2s, &\text{if } 0<s<1 .
\end{cases} $$
