There is exactly one line perpendicular to a given line l that passes through a point P not on l in Euclidean geometry. However, in the geometry of the sphere there are an uncountable number of lines through P that are perpendicular to l. This is because between any two points on the sphere, there are uncountably many other points, and each point defines a line through P that is perpendicular to l.

1. (2.) In Euclidean geometry, given a line l and a point P not on the line, there is exactly one line
through P that is perpendicular to l. Discuss the number of lines through P that are perpendicular
to l in the geometry of the sphere. Include a picture with your explanation.
Solution
Let us assume that we are in R2.
Draw a square containing the point P. Consider a vertex V of the square. By Euclid's axioms,
there is a line passing through the vertex V and the point P.
Next consider a vertex T adjacent to V.Again, we can find a line passing through T and V.
Between any two points in R2 there exists another point P1 on the line by virtue of denseness of
reals.
Again join a line between P1 and V.
As there are uncountable real numbers on any given interval on R, so there are uncountably
many points between P1 and V. Hence, you can draw uncountably many lines passing through P.
down vote
Let us assume that we are in R2.
Draw a square containing the point P. Consider a vertex V of the square. By Euclid's axioms,
there is a line passing through the vertex V and the point P.
Next consider a vertex T adjacent to V.Again, we can find a line passing through T and V.
Between any two points in R2 there exists another point P1 on the line by virtue of denseness of
reals.
Again join a line between P1 and V.
As there are uncountable real numbers on any given interval on R, so there are uncountably
many points between P1 and V. Hence, you can draw uncountably many lines passing through
P.