Finding distribution of $y$ intercept given by line generated by two uniformly chosen points in $[0,1]^2$ The method I'm using seems super inefficient. What I did was define $4$ RVs namely $X_1,X_2,Y_1,Y_2$ and thus my two uniformly random points are $(X_1,Y_1),(X_2,Y_2)$ and hence the $y$ intercept is $-X_1(Y_2-Y_1)/(X_2-X_1)+Y_1$ but them I'll have to do Jacobian tranform multiple times. 
Is there any simpler method I seem to be missing?

Partial answer: Proofs that (A) $y$-intercept $\in [0,1]$ with probability $\frac12$, and (B) it is uniform within $[0,1]$.
For shorthand, write $E=$ the event that $y$-intercept $\in [0,1]$.
Easy proof of (A): Given any two points $P_1, P_2 \in [0,1]^2$, the line $L$ through them hits a corner of the square with probability $0$.  Therefore, $L$ passes through exactly two sides of the square with probability $1$.  Now rotate the square by $90^\circ, 180^\circ, 270^\circ$.  The $4$ rotations are symmetric, and of the four lines exactly two of them passes through the left side of the square, i.e. has $y$-intercept $\in [0,1]$.  Thus $P(E) = \frac12$.
Harder proof of (A) and (B): Let $W = \max(X_1, X_2) =$ the $x$-coordinate of the point that is further to the right.  We call this point $P$ and the other point $Q$.  Conditioned on $W=w$, the point $Q$ is uniform in the rectangle $R$ with corners $(0,0), (0,1), (w,0), (w,1)$, i.e. the rectangle to the left of $x=w$.  Meanwhile, the $y$-intercept $\in [0,1]$ iff $Q$ lies in the triangle $T$ formed by $P, (0,0), (0,1)$.  This triangle $T$ is exactly $\frac12$ of rectangle $R$, so the conditional probability $P(E \mid W=w) = P(Q\in T \mid Q \in R) = \frac12$.  Since this is true for any $w$, the unconditioned probability $P(E)$ is also $\frac12$, regardless of the distribution of $W$.
More generally, the $y$-intercept $\in [0, c]$ (where $c\le 1$) iff $Q \in$ the smaller triangle $Z$ formed by $P, (0,0), (0,c)$.  This triangle $Z$'s area is exactly $c$ of the area of $T$, or equivalently, ${c \over 2}$ of the area of $R$.  This shows uniform distribution within $[0,1]$.
