Griffths buffons needle

Griffths buffons needle

I've seen a lot of other proofs online about buffons needle, and I
understand how they work, but I'm very confused about how Griffiths did
this.
I just can't visualize it.
Needle Length L on panels Length L. What's the probability that if dropped
it will cross a line?
He asks to refer to the previous question which goes as
"We consider a broken car speedometer, but we're interested in the x
coordinate of the speedometer point (The projection on the horizontal
line)"
Now all the questions are easy but Could you explain his answer?
"Suppose the eye end lands a distance y up from the line $ (0<y<L)$ and
let x be the projection along the same direction $(-L<x<L)$. The needle
crosses the line above if $(y+x>L) or (x>L-y)$, and it crosses when
$(y+x<0) or (x<-y)$. So the probability of the needle crossing the line is
( $$ P(y)=\int_{-l}^{-y} \rho(x)dx + \int_{l-y}^{l} \rho(x)dx=\frac1\pi
\{\int_{-l}^{-y} \frac1{\sqrt{l^2-x^2}} dx + \int_{l-y}^{l}
\frac1{\sqrt{l^2-x^2}} dx\} $$ And so on until I get $ \frac2\pi $
I understand the math. I don't understand how he got the boundary terms
(l-y) or (y) I can picture what if y was half way (l/2) and x was the full
extent (l). Then I picture a needle parrallel to the boundary, but not
crossing any lines??