Perpendicular proof (theorem)

1. Mar 31, 2014

Lee33

1. The problem statement, all variables and given/known data
Given a line $l$ and a point $B\in l$ in a protractor geometry, there exists a unique line $l'$ that contains $B$ such that $l\perp l'.$

2. Relevant equations

None

3. The attempt at a solution

I am not sure how to prove uniqueness or existence in this theorem.

We say lines $l$ and $m$ in a protractor geometry are perpendicular, denoted $l\perp m$, if $l\cup m$ contains a right angle.

2. Dec 15, 2014

PcumP_Ravenclaw

3. The attempt at a solution

Understanding the problem:

B is a unique point on the line $l$. A line is made up of many points. A unique perpendicular line $l'$ means a line that does not contain points that are in line $l$ except the point B. So we try to prove that a line $l$ is perpendicular to another line $l'$ then they should have only one point in common which is B. I can only prove with examples so

Take a line in the form y = mx + c then select a point B (any point) then draw the perpendicular line at the point B. By inspection you will see that the two lines only intersect at the point B. They will never intersect anywhere else so the two lines are unique except at the point B.

3. Dec 15, 2014

Staff: Mentor

We need to show that the perpendicular is unique, not the point B.

The usual style for proving uniqueness is to assume that opposite of the conclusion; i.e., that the perpendiculars are not unique, and show that this assumption leads to a contradiction. The contradiction means that the assumption must have been incorrect, so you are left with a single perpendicular.

Given a line L and a point B on L, assume that distinct lines L1 and L2 go through B and are perpendicular to L. Can you show that this assumption leads to a contradiction? If so, you will have proved the statement by contradiction.

4. Dec 15, 2014

Ray Vickson

There are two things to prove here:
(1) There exists at least one perpendicular of the type you describe; and
(2) There does not exist more than one such perpendicular.

If I were doing the question, I would begin by trying to show (1).