# How many regions do $n$ lines divide the plane into? [duplicate]

Suppose you draw $n \ge 0$ distinct lines in the plane, one after another, none of the lines parallel to any other and no three lines intersecting at a common point. The plane will, as a result, be divided into how many different regions $L_n$? Find an expression for $L_n$ in terms of $L_{n-1}$, solve it explicitly, and indicate what is $L_{10}$.

I have tried to come up with a solution but cannot. A little guidance would be very helpful.

## marked as duplicate by Joel Reyes Noche, user147263, Rebecca J. Stones, Najib Idrissi, Peter TaylorMar 12 '15 at 10:47

• Please give your questions more descriptive titles. People who are searching the site need to be able to see what a question is about from its title. – Zev Chonoles Sep 28 '13 at 18:51
• I feel like a similar question has been asked before but don't know where I might find the question. Does anyone know if this is a duplicate? – Dan Rust Sep 28 '13 at 18:54
• math.stackexchange.com/questions/339750/… – Magdiragdag Sep 28 '13 at 18:57

Each new line intersects other $n$ lines in one point and divides each previous region of space into two regions. Therefore each new line adds new $n+1$ region and so we have: $$L_{n+1}=L_n+n+1$$ with $L_1=2$.
Indeed you can see that $L_n=1+\frac{n(n+1)}{2}$ and hence $L_{10}=56$.
HINT: If you already have $n$ lines and add an $(n+1)$-st line, the new line crosses each of the original $n$ lines. Those intersections divide the new line into $k+1$ segments (including the two unbounded segments). Show that each of those segments divides one old region into two new ones. This gives you a recurrence expressing $L_{n+1}$ in terms of $L_n$, and it’s an easy recurrence to solve for a closed form.