Question

using induction

8. A set of n lines are drawn in the plane. No three lines meet at a common point. No two lines are na +n parallel. Then these lines divide the plane into 1 regions.

Answer 1

We can prove it by induction :

Proof: First, if n = 0, we note that if zero lines are drawn on the plane, there is a single region: thew wholeplane,

so the statement is true if n = 0

since
02 0+1-1

Next, assume that no matter how you draw k lines on the plane, consistent with the conditions oft he problem,

that there are exactly
2+k+2
regions formed for n=k. Consider a similar configuration of n=k + 1 lines. If we choose one of them and eliminate it, there will be
2+k+2
regions. When we look at the line we temporarily eliminated, since it is not parallel to any of the lines, it must intersect all of them: that makes k intersections.

None of these intersections are at the same point on the new line, or otherwise there would be three lines

intersecting at a point, which is not allowed according to the conditions of the problem.

The points of intersection thus divide the new line into k + 1 segments, each of which lies in a

different one of the
2+k+2
regions formed by the original k lines. That means that each of these k + 1 segments divides its region into two, so the addition of the (k + 1)st line adds k + 1 regions. Thus there are now:
+1
regions in the new configuration.

This is equivalent to

12 +k + 2 2k + 2

12 +k +11 2k 2

$=\frac{(k^2+2k+1)+(k+1)+2}{2}$$=\frac{(k+1)^2+(k+1)+2}{2}$regions, which is what we needed to show.